Algebraic Expressions and Identities

Multiplication and division do not separate the terms of an algebraic expression. Only the parts separated by '+' or '-' sign are counted as terms.

$5p × q × r^2$ is a product of factors, so it has only one term.

Hence, $\bm{5p × q × r^2}$ is a monomial.

Multiplication and division do not separate the terms of an algebraic expression. Only the parts separated by '+' or '-' sign are counted as terms.

$3x^2 \times y \div 2z$ involves only multiplication and division, so it has only one term.

Hence, $\bm{3x^2 \times y \div 2z}$ is a monomial.

Multiplication and division do not separate the terms of an algebraic expression. Only the parts separated by '+' or '-' sign are counted as terms.

$-3 + 7x^2$ has two terms, namely -3 and $7x^2$.

Hence, $\bm{-3 + 7x^2}$ is a binomial.

Multiplication and division do not separate the terms of an algebraic expression. Only the parts separated by '+' or '-' sign are counted as terms.

$\frac{5a^2 - 3b^2 + c}{2} = \frac{5a^2}{2} - \frac{3b^2}{2} + \frac{c}{2}$, which has three terms.

Hence, $\bm{\frac{5a^2 - 3b^2 + c}{2}}$ is a trinomial.

Multiplication and division do not separate the terms of an algebraic expression. Only the parts separated by '+' or '-' sign are counted as terms.

$7x^5 - \frac{3x}{y}$ has two terms, namely $7x^5$ and $-\frac{3x}{y}$.

Hence, $\bm{7x^5 - \frac{3x}{y}}$ is a binomial.

Multiplication and division do not separate the terms of an algebraic expression. Only the parts separated by '+' or '-' sign are counted as terms.

$5p \div 3q - 3p^2 \times q^2$ has two terms, namely $5p \div 3q$ and $3p^2 \times q^2$.

Hence, $\bm{5p \div 3q - 3p^2 \times q^2}$ is a binomial.

An algebraic expression is a polynomial if the powers of the variables in each term are non-negative integers. The degree is the greatest sum of the powers of the variables in a term.

(i) $\frac{2}{5}x^4 - \sqrt{3}x^2 + 5x - 1$

The powers of $x$ in the terms are 4, 2, 1 and 0, which are all non-negative integers.

So, it is a polynomial and the greatest power is 4.

Hence, it is a polynomial of degree 4.

(ii) $7x^3 - \frac{3}{x^2} + \sqrt{5}$

Here $\frac{3}{x^2} = 3x^{-2}$, in which the power of $x$ is -2, a negative integer.

Hence, it is not a polynomial.

(iii) $4a^3b^2 - 3ab^4 + 5ab + \frac{2}{3}$

The powers of $a$ and $b$ in each term are non-negative integers, so it is a polynomial.

The sums of the powers of the terms are 3 + 2 = 5, 1 + 4 = 5, 1 + 1 = 2 and 0.

The greatest sum is 5.

Hence, it is a polynomial of degree 5.

(iv) $2x^2y - \frac{3}{xy} + 5y^3 + \sqrt{3}$

Here $\frac{3}{xy} = 3x^{-1}y^{-1}$, in which the powers of $x$ and $y$ are negative integers.

Hence, it is not a polynomial.

Write each expression in a separate row, keeping the like terms one below the other, and then add them.

$$\begin{array}{rrrr} & ab & - bc & \\ & & + bc & - ca \\ + & - ab & & + ca \\ \hline & 0 & & \\ \end{array}$$

Hence, the required sum = 0.

Write each expression in a separate row, keeping the like terms one below the other, and then add them.

$$\begin{array}{rrrr} & 5p^2q^2 & + 4pq & + 7 \\ + & - 2p^2q^2 & + 9pq & + 3 \\ \hline & 3p^2q^2 & + 13pq & + 10 \\ \end{array}$$

Hence, the required sum = $\bm{3p^2q^2 + 13pq + 10}$.

Write each expression in a separate row, keeping the like terms one below the other, and then add them.

$$\begin{array}{rrrrrrr} & l^2 & + m^2 & + n^2 & & & \\ & & & & lm & + mn & \\ & & & & & + mn & + nl \\ + & & & & + lm & & + nl \\ \hline & l^2 & + m^2 & + n^2 & + 2lm & + 2mn & + 2nl \\ \end{array}$$

The sum can also be written as $l^2 + m^2 + n^2 + 2(lm + mn + nl)$.

Hence, the required sum = $\bm{l^2 + m^2 + n^2 + 2(lm + mn + nl)}$.

Write each expression in a separate row, keeping the like terms one below the other, and then add them.

$$\begin{array}{rrrrr} & 4x^3 & - 7x^2 & & + 9 \\ & & + 3x^2 & - 5x & + 4 \\ & 7x^3 & & - 11x & + 1 \\ + & & + 6x^2 & - 13x & \\ \hline & 11x^3 & + 2x^2 & - 29x & + 14 \\ \end{array}$$

Hence, the required sum = $\bm{11x^3 + 2x^2 - 29x + 14}$.

Subtracting,

$$\begin{array}{rrrrr} & 14a & - 5ab & + 7b & - 5 \\ & 8a & + 3ab & - 2b & + 7 \\ &- &- &+ &- \\ \hline & 6a & - 8ab & + 9b & - 12 \\ \end{array}$$

Hence, the required difference = $\bm{6a - 8ab + 9b - 12}$.

Subtracting,

$$\begin{array}{rrrrr} & 12xy & - 3yz & - 4zx & + 5xyz \\ & 8xy & + 4yz & + 5zx & \\ &- &- &- & \\ \hline & 4xy & - 7yz & - 9zx & + 5xyz \\ \end{array}$$

Hence, the required difference = $\bm{4xy - 7yz - 9zx + 5xyz}$.

Subtracting,

$$\begin{array}{rrrrrrr} & 5p^2q & - 2pq^2 & + 5pq & - 3p & - 11q & + 18 \\ & 4p^2q & + 5pq^2 & - 3pq & - 8p & + 7q & - 10 \\ &- &- &+ &+ &- &+ \\ \hline & p^2q & - 7pq^2 & + 8pq & + 5p & - 18q & + 28 \\ \end{array}$$

Hence, the required difference = $\bm{p^2q + 8pq - 7pq^2 + 5p - 18q + 28}$.

First we find the sum of the two given expressions by the column method.

$$\begin{array}{rrrrr} & 3x^2 & + 5xy & + 7y^2 & + 3 \\ + & 2x^2 & - 4xy & - 3y^2 & + 7 \\ \hline & 5x^2 & + xy & + 4y^2 & + 10 \\ \end{array}$$

Now we subtract this sum from $9x^2 - 8xy + 11y^2$.

$$\begin{array}{rrrrr} & 9x^2 & - 8xy & + 11y^2 & \\ & 5x^2 & + xy & + 4y^2 & + 10 \\ &- &- &- &- \\ \hline & 4x^2 & - 9xy & + 7y^2 & - 10 \\ \end{array}$$

Hence, the required difference = $\bm{4x^2 - 9xy + 7y^2 - 10}$.

Let the required expression be A.

According to the question,

⇒ $(3a^2 - 5ab - 2b^2 - 3) - A = 5a^2 - 7ab - 3b^2 + 3a$

⇒ $A = (3a^2 - 5ab - 2b^2 - 3) - (5a^2 - 7ab - 3b^2 + 3a)$

Subtracting,

$$\begin{array}{rrrrrr} & 3a^2 & - 5ab & - 2b^2 & & - 3 \\ & 5a^2 & - 7ab & - 3b^2 & + 3a & \\ &- &+ &+ &- & \\ \hline & - 2a^2 & + 2ab & + b^2 & - 3a & - 3 \\ \end{array}$$

Hence, $\bm{-2a^2 + 2ab + b^2 - 3a - 3}$ must be subtracted.

As the perimeter of a triangle is the sum of its three sides,

Third side = Perimeter - Sum of the other two sides

First we find the sum of the two given sides by the column method.

$$\begin{array}{rrrr} & p^2 & + 2p & - 1 \\ + & 3p^2 & - 6p & + 3 \\ \hline & 4p^2 & - 4p & + 2 \\ \end{array}$$

Now we subtract this sum from the perimeter $7p^2 - 5p + 11$.

$$\begin{array}{rrrr} & 7p^2 & - 5p & + 11 \\ & 4p^2 & - 4p & + 2 \\ &- &+ &- \\ \hline & 3p^2 & - p & + 9 \\ \end{array}$$

Hence, the third side of the triangle = $\bm{3p^2 - p + 9}$.

Product of two or more monomials = (product of their numerical coefficients) × (product of their literal coefficients)

$$\begin{aligned}\Rightarrow 4x^3 \times (-3xy) \\[1em] \Rightarrow (4 \times (-3)) \times (x^3 \times x) \times y \\[1em] \Rightarrow -12x^{(3+1)}y \\[1em] \Rightarrow -12x^4y\end{aligned}$$

Hence, the required product = $\bm{-12x^4y}$.

Product of two or more monomials = (product of their numerical coefficients) × (product of their literal coefficients)

$$\begin{aligned}\Rightarrow 2xyz \times 0 \\[1em] \Rightarrow (2 \times 0) \times xyz \\[1em] \Rightarrow 0\end{aligned}$$

Hence, the required product = 0.

Product of two or more monomials = (product of their numerical coefficients) × (product of their literal coefficients)

$$\begin{aligned}\Rightarrow -\frac{2}{3}p^2q \times \frac{3}{4}pq^2 \times 5pqr \\[1em] \Rightarrow \Big(-\frac{2}{3} \times \frac{3}{4} \times 5\Big) \times p^{2+1+1} \times q^{1+2+1} \times r \\[1em] \Rightarrow -\frac{2 \times 3 \times 5}{3 \times 4} \times p^4q^4r \\[1em] \Rightarrow -\frac{30}{12}p^4q^4r \\[1em] \Rightarrow -\frac{5}{2}p^4q^4r\end{aligned}$$

Hence, the required product = $\bm{-\frac{5}{2}p^4q^4r}$.

Product of two or more monomials = (product of their numerical coefficients) × (product of their literal coefficients)

$$\begin{aligned}\Rightarrow -\frac{1}{2}x^2 \times \Big(-\frac{3}{5}xy\Big) \times \frac{2}{3}yz \times \frac{5}{7}xyz \\[1em] \Rightarrow \Big(-\frac{1}{2} \times -\frac{3}{5} \times \frac{2}{3} \times \frac{5}{7}\Big) \times x^{2+1+1} \times y^{1+1+1} \times z^{1+1} \\[1em] \Rightarrow \frac{1 \times 3 \times 2 \times 5}{2 \times 5 \times 3 \times 7} \times x^4y^3z^2 \\[1em] \Rightarrow \frac{30}{210}x^4y^3z^2 \\[1em] \Rightarrow \frac{1}{7}x^4y^3z^2\end{aligned}$$

Hence, the required product = $\bm{\frac{1}{7}x^4y^3z^2}$.

Area of rectangle = length × breadth

$$\begin{aligned}\Rightarrow p^2q \times pq^2 \\[1em] \Rightarrow (p^2 \times p) \times (q \times q^2) \\[1em] \Rightarrow p^{(2+1)}q^{(1+2)} \\[1em] \Rightarrow p^3q^3\end{aligned}$$

Hence, the area of the rectangle = $\bm{p^3q^3}$.

Area of rectangle = length × breadth

$$\begin{aligned}\Rightarrow 5xy \times 7xy^2 \\[1em] \Rightarrow (5 \times 7) \times (x \times x) \times (y \times y^2) \\[1em] \Rightarrow 35x^{(1+1)}y^{(1+2)} \\[1em] \Rightarrow 35x^2y^3\end{aligned}$$

Hence, the area of the rectangle = $\bm{35x^2y^3}$.

Volume of a rectangular box = length × breadth × height

$$\begin{aligned}\Rightarrow 5ab \times 3a^2b \times 7a^4b^2 \\[1em] \Rightarrow (5 \times 3 \times 7) \times (a \times a^2 \times a^4) \times (b \times b \times b^2) \\[1em] \Rightarrow 105a^{(1+2+4)}b^{(1+1+2)} \\[1em] \Rightarrow 105a^7b^4\end{aligned}$$

Hence, the volume of the rectangular box = $\bm{105a^7b^4}$.

Volume of a rectangular box = length × breadth × height

$$\begin{aligned}\Rightarrow 2pq \times 4q^2 \times 8rp \\[1em] \Rightarrow (2 \times 4 \times 8) \times (p \times p) \times (q \times q^2) \times r \\[1em] \Rightarrow 64p^{(1+1)}q^{(1+2)}r \\[1em] \Rightarrow 64p^2q^3r\end{aligned}$$

Hence, the volume of the rectangular box = $\bm{64p^2q^3r}$.

To multiply a polynomial by a monomial, we multiply each term of the polynomial by the monomial.

$$\begin{aligned}\Rightarrow (3x - 5y + 7z) \times (-3xyz) \\[1em] \Rightarrow 3x \times (-3xyz) - 5y \times (-3xyz) + 7z \times (-3xyz) \\[1em] \Rightarrow -9x^{(1+1)}yz + 15xy^{(1+1)}z - 21xyz^{(1+1)} \\[1em] \Rightarrow -9x^2yz + 15xy^2z - 21xyz^2\end{aligned}$$

Hence, the required product = $\bm{-9x^2yz + 15xy^2z - 21xyz^2}$.

To multiply a polynomial by a monomial, we multiply each term of the polynomial by the monomial.

$$\begin{aligned}\Rightarrow (2p^2 - 3pq + 5q^2 + 5) \times (-2pq) \\[1em] \Rightarrow 2p^2 \times (-2pq) - 3pq \times (-2pq) + 5q^2 \times (-2pq) + 5 \times (-2pq) \\[1em] \Rightarrow -4p^{(2+1)}q + 6p^{(1+1)}q^{(1+1)} - 10pq^{(2+1)} - 10pq \\[1em] \Rightarrow -4p^3q + 6p^2q^2 - 10pq^3 - 10pq\end{aligned}$$

Hence, the required product = $\bm{-4p^3q + 6p^2q^2 - 10pq^3 - 10pq}$.

To multiply a polynomial by a monomial, we multiply each term of the polynomial by the monomial.

$$\begin{aligned}\Rightarrow \Big(\frac{2}{3}a^2b - \frac{4}{5}ab^2 + \frac{2}{7}ab + 3\Big) \times 35ab \\[1em] \Rightarrow \frac{2}{3}a^2b \times 35ab - \frac{4}{5}ab^2 \times 35ab + \frac{2}{7}ab \times 35ab + 3 \times 35ab \\[1em] \Rightarrow \frac{2 \times 35}{3}a^3b^2 - \frac{4 \times 35}{5}a^2b^3 + \frac{2 \times 35}{7}a^2b^2 + 105ab \\[1em] \Rightarrow \frac{70}{3}a^3b^2 - 28a^2b^3 + 10a^2b^2 + 105ab\end{aligned}$$

Hence, the required product = $\bm{\frac{70}{3}a^3b^2 - 28a^2b^3 + 10a^2b^2 + 105ab}$.

To multiply a polynomial by a monomial, we multiply each term of the polynomial by the monomial.

$$\begin{aligned}\Rightarrow (4x^2 - 10xy + 7y^2 - 8x + 4y + 3) \times 3xy \\[1em] \Rightarrow 4x^2 \times 3xy - 10xy \times 3xy + 7y^2 \times 3xy - 8x \times 3xy + 4y \times 3xy + 3 \times 3xy \\[1em] \Rightarrow 12x^3y - 30x^2y^2 + 21xy^3 - 24x^2y + 12xy^2 + 9xy\end{aligned}$$

Hence, the required product =

$\bm{12x^3y - 30x^2y^2 + 21xy^3 - 24x^2y + 12xy^2 + 9xy}$.

First we simplify the expression by multiplying each term of the polynomial by the monomial.

$$\begin{aligned}\Rightarrow x^2(3 - 2x + x^2) \\[1em] \Rightarrow x^2 \times 3 - x^2 \times 2x + x^2 \times x^2 \\[1em] \Rightarrow 3x^2 - 2x^3 + x^4 \\[1em] \Rightarrow x^4 - 2x^3 + 3x^2\end{aligned}$$

For $x = 1$, the value of $x^4 - 2x^3 + 3x^2$ is

$$\begin{aligned}\Rightarrow (1)^4 - 2(1)^3 + 3(1)^2 \\[1em] \Rightarrow 1 - 2 + 3 \\[1em] \Rightarrow 2\end{aligned}$$

For $x = -1$, the value of $x^4 - 2x^3 + 3x^2$ is

$$\begin{aligned}\Rightarrow (-1)^4 - 2(-1)^3 + 3(-1)^2 \\[1em] \Rightarrow 1 + 2 + 3 \\[1em] \Rightarrow 6\end{aligned}$$

For $x = \frac{2}{3}$, the value of $x^4 - 2x^3 + 3x^2$ is

$$\begin{aligned}\Rightarrow \Big(\frac{2}{3}\Big)^4 - 2\Big(\frac{2}{3}\Big)^3 + 3\Big(\frac{2}{3}\Big)^2 \\[1em] \Rightarrow \frac{16}{81} - 2 \times \frac{8}{27} + 3 \times \frac{4}{9} \\[1em] \Rightarrow \frac{16}{81} - \frac{16}{27} + \frac{4}{3} \\[1em] \Rightarrow \frac{16 - 48 + 108}{81} \\[1em] \Rightarrow \frac{76}{81}\end{aligned}$$

For $x = -\frac{1}{2}$, the value of $x^4 - 2x^3 + 3x^2$ is

$$\begin{aligned}\Rightarrow \Big(-\frac{1}{2}\Big)^4 - 2\Big(-\frac{1}{2}\Big)^3 + 3\Big(-\frac{1}{2}\Big)^2 \\[1em] \Rightarrow \frac{1}{16} + 2 \times \frac{1}{8} + 3 \times \frac{1}{4} \\[1em] \Rightarrow \frac{1}{16} + \frac{1}{4} + \frac{3}{4} \\[1em] \Rightarrow \frac{1 + 4 + 12}{16} \\[1em] \Rightarrow \frac{17}{16}\end{aligned}$$

Hence, $\bm{x^2(3 - 2x + x^2) = x^4 - 2x^3 + 3x^2}$ and its values are 2, 6, $\bm{\frac{76}{81}}$ and $\bm{\frac{17}{16}}$ respectively.

First we simplify the expression by multiplying each polynomial by the monomial outside it.

$$\begin{aligned}\Rightarrow 5xy(3x + 4y - 7) - 3y(xy - x^2 + 9) - 8 \\[1em] \Rightarrow 15x^2y + 20xy^2 - 35xy - 3xy^2 + 3x^2y - 27y - 8 \\[1em] \Rightarrow (15x^2y + 3x^2y) + (20xy^2 - 3xy^2) - 35xy - 27y - 8 \\[1em] \Rightarrow 18x^2y + 17xy^2 - 35xy - 27y - 8\end{aligned}$$

For $x = 2$, $y = -1$, the value of $18x^2y + 17xy^2 - 35xy - 27y - 8$ is

$$\begin{aligned}\Rightarrow 18(2)^2(-1) + 17(2)(-1)^2 - 35(2)(-1) - 27(-1) - 8 \\[1em] \Rightarrow 18 \times 4 \times (-1) + 17 \times 2 \times 1 + 70 + 27 - 8 \\[1em] \Rightarrow -72 + 34 + 70 + 27 - 8 \\[1em] \Rightarrow 51\end{aligned}$$

Hence, the simplified expression = $\bm{18x^2y + 17xy^2 - 35xy - 27y - 8}$ and its value = 51.

First we simplify the given expressions and then collect the like terms.

$$\begin{aligned}\Rightarrow 4p(2 - p^2) + 8p^3 - 3p \\[1em] \Rightarrow 8p - 4p^3 + 8p^3 - 3p \\[1em] \Rightarrow (-4p^3 + 8p^3) + (8p - 3p) \\[1em] \Rightarrow 4p^3 + 5p\end{aligned}$$

Hence, the required sum = $\bm{4p^3 + 5p}$.

First we simplify the given expressions and then collect the like terms.

$$\begin{aligned}\Rightarrow 7xy(8x + 2y - 3) + 4xy^2(3y - 7x + 8) \\[1em] \Rightarrow 56x^2y + 14xy^2 - 21xy + 12xy^3 - 28x^2y^2 + 32xy^2 \\[1em] \Rightarrow 12xy^3 - 28x^2y^2 + 56x^2y + (14xy^2 + 32xy^2) - 21xy \\[1em] \Rightarrow 12xy^3 - 28x^2y^2 + 56x^2y + 46xy^2 - 21xy\end{aligned}$$

Hence, the required sum = $\bm{12xy^3 - 28x^2y^2 + 56x^2y + 46xy^2 - 21xy}$.

First we simplify both the expressions and then subtract.

Simplifying first expression:

$$\begin{aligned}\Rightarrow 6x(x - y + z) - 3y(x + y - z) \\[1em] \Rightarrow 6x^2 - 6xy + 6zx - 3xy - 3y^2 + 3yz \\[1em] \Rightarrow 6x^2 - 9xy - 3y^2 + 3yz + 6zx\end{aligned}$$

Simplifying second expression:

$$\begin{aligned}\Rightarrow 2z(-x + y + z) \\[1em] \Rightarrow -2zx + 2yz + 2z^2\end{aligned}$$

Subtracting first expression from second expression:

$$\begin{aligned}\Rightarrow (-2zx + 2yz + 2z^2) - (6x^2 - 9xy - 3y^2 + 3yz + 6zx) \\[1em] \Rightarrow -2zx + 2yz + 2z^2 - 6x^2 + 9xy + 3y^2 - 3yz - 6zx \\[1em] \Rightarrow -6x^2 + 3y^2 + 2z^2 + 9xy + (2yz - 3yz) + (-2zx - 6zx) \\[1em] \Rightarrow -6x^2 + 3y^2 + 2z^2 + 9xy - yz - 8zx\end{aligned}$$

Hence, the required difference = $\bm{-6x^2 + 3y^2 + 2z^2 + 9xy - yz - 8zx}$.

First we simplify both the expressions and then subtract.

Simplifying first expression:

$$\begin{aligned}\Rightarrow 7xy(x^2 - 2xy + 3y^2) - 8x(x^2y - 4xy + 7xy^2) \\[1em] \Rightarrow 7x^3y - 14x^2y^2 + 21xy^3 - 8x^3y + 32x^2y - 56x^2y^2 \\[1em] \Rightarrow (7x^3y - 8x^3y) + (-14x^2y^2 - 56x^2y^2) + 21xy^3 + 32x^2y \\[1em] \Rightarrow -x^3y - 70x^2y^2 + 21xy^3 + 32x^2y\end{aligned}$$

Simplifying second expression:

$$\begin{aligned}\Rightarrow 3y(4x^2y - 5xy + 8xy^2) \\[1em] \Rightarrow 12x^2y^2 - 15xy^2 + 24xy^3\end{aligned}$$

Subtracting first expression from second expression:

$$\begin{aligned}\Rightarrow (12x^2y^2 - 15xy^2 + 24xy^3) - (-x^3y - 70x^2y^2 + 21xy^3 + 32x^2y) \\[1em] \Rightarrow 12x^2y^2 - 15xy^2 + 24xy^3 + x^3y + 70x^2y^2 - 21xy^3 - 32x^2y \\[1em] \Rightarrow (12x^2y^2 + 70x^2y^2) + (24xy^3 - 21xy^3) + x^3y - 15xy^2 - 32x^2y \\[1em] \Rightarrow 82x^2y^2 + 3xy^3 + x^3y - 15xy^2 - 32x^2y\end{aligned}$$

Hence, the required difference = $\bm{82x^2y^2 + 3xy^3 + x^3y - 15xy^2 - 32x^2y}$.

Solving,

$$\begin{array}{rrrr} & & 5x & - 2 \\ \times & & 3x & + 4 \\ \hline & 15x^2 & - 6x & \\ & & + 20x & - 8 \\ \hline & 15x^2 & + 14x & - 8 \\ \end{array}$$

Hence, the required product = $\bm{15x^2 + 14x - 8}$.

Solving,

$$\begin{array}{rrrr} & & ax & + b \\ \times & & cx & + d \\ \hline & acx^2 & + bcx & \\ & & + adx & + bd \\ \hline & acx^2 & + (ad + bc)x & + bd \\ \end{array}$$

Hence, the required product = $\bm{acx^2 + (ad + bc)x + bd}$.

Solving,

Arranging the terms of the second polynomial in descending powers of $p$, we get $-3p + 2$.

$$\begin{array}{rrrr} & & 4p & - 7 \\ \times & & -3p & + 2 \\ \hline & -12p^2 & + 21p & \\ & & + 8p & - 14 \\ \hline & -12p^2 & + 29p & - 14 \\ \end{array}$$

Hence, the required product = $\bm{-12p^2 + 29p - 14}$.

Solving,

$$\begin{array}{rrrrr} & & 2x^2 & & + 3 \\ \times & & & 3x & - 5 \\ \hline & 6x^3 & & + 9x & \\ & & - 10x^2 & & - 15 \\ \hline & 6x^3 & - 10x^2 & + 9x & - 15 \\ \end{array}$$

Hence, the required product = $\bm{6x^3 - 10x^2 + 9x - 15}$.

Solving,

$$\begin{array}{rrrr} & & 1.5a & - 2.5b \\ \times & & 1.5a & + 2.5b \\ \hline & 2.25a^2 & - 3.75ab & \\ & & + 3.75ab & - 6.25b^2 \\ \hline & 2.25a^2 & & - 6.25b^2 \\ \end{array}$$

Hence, the required product = $\bm{2.25a^2 - 6.25b^2}$.

Solving,

First we simplify the second polynomial as $7\Big(p^2 - \frac{3}{4}q^2\Big) = 7p^2 - \frac{21}{4}q^2$.

$$\begin{array}{rrrr} & & \frac{3}{7}p^2 & + 4q^2 \\ \times & & 7p^2 & - \frac{21}{4}q^2 \\ \hline & 3p^4 & + 28p^2q^2 & \\ & & - \frac{9}{4}p^2q^2 & - 21q^4 \\ \hline & 3p^4 & + \frac{103}{4}p^2q^2 & - 21q^4 \\ \end{array}$$

Hence, the required product = $\bm{3p^4 + \frac{103}{4}p^2q^2 - 21q^4}$.

Solving,

$$\begin{array}{rrrrrr} & & & x & - 2y & + 3 \\ \times & & & & x & + 2y \\ \hline & x^2 & - 2xy & & + 3x & \\ & & + 2xy & - 4y^2 & & + 6y \\ \hline & x^2 & & - 4y^2 & + 3x & + 6y \\ \end{array}$$

Hence, the required product = $\bm{x^2 - 4y^2 + 3x + 6y}$.

Solving,

Arranging the terms of the first polynomial in descending powers of $x$, we get $2x^2 - 5x + 3$.

$$\begin{array}{rrrrr} & & 2x^2 & - 5x & + 3 \\ \times & & & 4x & - 5 \\ \hline & 8x^3 & - 20x^2 & + 12x & \\ & & - 10x^2 & + 25x & - 15 \\ \hline & 8x^3 & - 30x^2 & + 37x & - 15 \\ \end{array}$$

Hence, the required product = $\bm{8x^3 - 30x^2 + 37x - 15}$.

Solving,

$$\begin{array}{rrrrrr} & & & 3x^2 & - 2x & - 1 \\ \times & & & 2x^2 & + x & - 5 \\ \hline & 6x^4 & - 4x^3 & - 2x^2 & & \\ & & + 3x^3 & - 2x^2 & - x & \\ & & & - 15x^2 & + 10x & + 5 \\ \hline & 6x^4 & - x^3 & - 19x^2 & + 9x & + 5 \\ \end{array}$$

Hence, the required product = $\bm{6x^4 - x^3 - 19x^2 + 9x + 5}$.

Solving,

Arranging the terms of both the polynomials in descending powers of $y$, we get $-5y^2 - 3y + 2$ and $3y^2 + 2y - 1$.

$$\begin{array}{rrrrrr} & & & -5y^2 & - 3y & + 2 \\ \times & & & 3y^2 & + 2y & - 1 \\ \hline & -15y^4 & - 9y^3 & + 6y^2 & & \\ & & - 10y^3 & - 6y^2 & + 4y & \\ & & & + 5y^2 & + 3y & - 2 \\ \hline & -15y^4 & - 19y^3 & + 5y^2 & + 7y & - 2 \\ \end{array}$$

Hence, the required product = $\bm{-15y^4 - 19y^3 + 5y^2 + 7y - 2}$.

First we find the product $(x^2 + 3)(x - 3)$ by the column method, in which we write the two polynomials one below the other, multiply the first polynomial by each term of the second polynomial and then add the like terms.

$$\begin{array}{rrrrr} & & x^2 & & + 3 \\ \times & & & x & - 3 \\ \hline & x^3 & & + 3x & \\ & & - 3x^2 & & - 9 \\ \hline & x^3 & - 3x^2 & + 3x & - 9 \\ \end{array}$$

Now, adding 9 to the product,

$$\begin{aligned}\Rightarrow x^3 - 3x^2 + 3x - 9 + 9\\[1em] \Rightarrow x^3 - 3x^2 + 3x\end{aligned}$$

Hence, $\bm{(x^2 + 3)(x - 3) + 9 = x^3 - 3x^2 + 3x}$.

Solving,

Taking the first two factors,

$$\begin{array}{rrrr} & & x & + 3 \\ \times & & x & - 3 \\ \hline & x^2 & + 3x & \\ & & - 3x & - 9 \\ \hline & x^2 & & - 9 \\ \end{array}$$

Taking the last two factors,

$$\begin{array}{rrrr} & & x & + 4 \\ \times & & x & - 4 \\ \hline & x^2 & + 4x & \\ & & - 4x & - 16 \\ \hline & x^2 & & - 16 \\ \end{array}$$

Now, multiplying the two results,

$$\begin{array}{rrrr} & & x^2 & - 9 \\ \times & & x^2 & - 16 \\ \hline & x^4 & - 9x^2 & \\ & & - 16x^2 & + 144 \\ \hline & x^4 & - 25x^2 & + 144 \\ \end{array}$$

Hence, $\bm{(x + 3)(x - 3)(x + 4)(x - 4) = x^4 - 25x^2 + 144}$.

Solving,

Taking the first two factors,

$$\begin{array}{rrrr} & & x & + 5 \\ \times & & x & + 6 \\ \hline & x^2 & + 5x & \\ & & + 6x & + 30 \\ \hline & x^2 & + 11x & + 30 \\ \end{array}$$

Now, multiplying this result by the third factor,

$$\begin{array}{rrrrr} & & x^2 & + 11x & + 30 \\ \times & & & x & + 7 \\ \hline & x^3 & + 11x^2 & + 30x & \\ & & + 7x^2 & + 77x & + 210 \\ \hline & x^3 & + 18x^2 & + 107x & + 210 \\ \end{array}$$

Hence, $\bm{(x + 5)(x + 6)(x + 7) = x^3 + 18x^2 + 107x + 210}$.

Solving $(p + q - 2r)(2p - q + r)$,

$$\begin{array}{rrrrrrr} & & & & p & + q & - 2r \\ \times & & & & 2p & - q & + r \\ \hline & 2p^2 & & & + 2pq & & - 4rp \\ & & - q^2 & & - pq & + 2qr & \\ & & & - 2r^2 & & + qr & + rp \\ \hline & 2p^2 & - q^2 & - 2r^2 & + pq & + 3qr & - 3rp \\ \end{array}$$

Now, subtracting $4qr$ from the product,

$$\begin{aligned}\Rightarrow (p + q - 2r)(2p - q + r) - 4qr\\[1em] \Rightarrow 2p^2 - q^2 - 2r^2 + pq + 3qr - 3rp - 4qr\\[1em] \Rightarrow 2p^2 - q^2 - 2r^2 + pq - qr - 3rp\end{aligned}$$

Hence, $\bm{(p + q - 2r)(2p - q + r) - 4qr = 2p^2 - q^2 - 2r^2 + pq - qr - 3rp}$.

Solving,

Taking the first product,

$$\begin{array}{rrrrr} & & & p & + q \\ \times & & & r & + s \\ \hline & pr & & + qr & \\ & & + ps & & + qs \\ \hline & pr & + ps & + qr & + qs \\ \end{array}$$

Taking the second product,

$$\begin{array}{rrrrr} & & & p & - q \\ \times & & & r & - s \\ \hline & pr & & - qr & \\ & & - ps & & + qs \\ \hline & pr & - ps & - qr & + qs \\ \end{array}$$

Now, adding the two products and subtracting $2(pr + qs)$,

$$\begin{aligned}\Rightarrow (pr + ps + qr + qs) + (pr - ps - qr + qs) - 2(pr + qs)\\[1em] \Rightarrow (pr + pr - 2pr) + (ps - ps) + (qr - qr) + (qs + qs - 2qs)\\[1em] \Rightarrow 0\end{aligned}$$

Hence, $\bm{(p + q)(r + s) + (p - q)(r - s) - 2(pr + qs) = 0}$.

Solving,

Taking the first product,

$$\begin{array}{rrrrrrr} & & & & x & + y & + z \\ \times & & & & x & - y & + z \\ \hline & x^2 & & & + xy & & + zx \\ & & - y^2 & & - xy & - yz & \\ & & & + z^2 & & + yz & + zx \\ \hline & x^2 & - y^2 & + z^2 & & & + 2zx \\ \end{array}$$

Taking the second product,

$$\begin{array}{rrrrrrr} & & & & x & + y & - z \\ \times & & & & -x & + y & + z \\ \hline & - x^2 & & & - xy & & + zx \\ & & + y^2 & & + xy & - yz & \\ & & & - z^2 & & + yz & + zx \\ \hline & - x^2 & + y^2 & - z^2 & & & + 2zx \\ \end{array}$$

Now, adding the two products and subtracting $4zx$,

$$\begin{aligned}\Rightarrow (x^2 - y^2 + z^2 + 2zx) + (-x^2 + y^2 - z^2 + 2zx) - 4zx\\[1em] \Rightarrow (x^2 - x^2) + (-y^2 + y^2) + (z^2 - z^2) + (2zx + 2zx - 4zx)\\[1em] \Rightarrow 0\end{aligned}$$

Hence, $\bm{(x + y + z)(x - y + z) + (x + y - z)(-x + y + z) - 4zx = 0}$.

Area of rectangle = length × breadth

We find the product by the column method, in which we write the two polynomials one below the other, multiply the first polynomial by each term of the second polynomial and then add the like terms.

$$\begin{array}{rrrrrr} & & & 5x^2 & + 25xy & + 4y^2 \\ \times & & & 2x^2 & - 2xy & + 3y^2 \\ \hline & 10x^4 & + 50x^3y & + 8x^2y^2 & & \\ & & - 10x^3y & - 50x^2y^2 & - 8xy^3 & \\ & & & + 15x^2y^2 & + 75xy^3 & + 12y^4 \\ \hline & 10x^4 & + 40x^3y & - 27x^2y^2 & + 67xy^3 & + 12y^4 \\ \end{array}$$

Hence, the area of the rectangle = $\bm{10x^4 + 40x^3y - 27x^2y^2 + 67xy^3 + 12y^4}$.

Solving,

$$\begin{aligned}\Rightarrow -39pq^2r^5 \div (-24p^3q^3r) \\[1em] \Rightarrow \frac{-39pq^2r^5}{-24p^3q^3r}\\[1em] \Rightarrow \frac{39}{24} \times \frac{pq^2r^5}{p^3q^3r}\\[1em] \Rightarrow \frac{13}{8} \times \frac{r^{5-1}}{p^{3-1}q^{3-2}}\\[1em] \Rightarrow \frac{13r^4}{8p^2q}\end{aligned}$$

Hence, the required quotient = $\bm{\frac{13r^4}{8p^2q}}$.

Solving,

$$\begin{aligned}\Rightarrow -\frac{3}{4}a^2b^3 \div \frac{6}{7}a^3b^2\\[1em] \Rightarrow -\frac{3}{4}a^2b^3 \times \frac{7}{6a^3b^2}\\[1em] \Rightarrow -\frac{3 \times 7}{4 \times 6} \times \frac{a^2b^3}{a^3b^2}\\[1em] \Rightarrow -\frac{21}{24} \times \frac{b^{3-2}}{a^{3-2}}\\[1em] \Rightarrow -\frac{7b}{8a}\end{aligned}$$

Hence, the required quotient = $\bm{-\frac{7b}{8a}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{9x^4 - 8x^3 - 12x + 3}{3x}\\[1em] \Rightarrow \frac{9x^4}{3x} - \frac{8x^3}{3x} - \frac{12x}{3x} + \frac{3}{3x}\\[1em] \Rightarrow 3x^{4-1} - \frac{8}{3}x^{3-1} - 4 + \frac{1}{x}\\[1em] \Rightarrow 3x^3 - \frac{8}{3}x^2 - 4 + \frac{1}{x}\end{aligned}$$

Hence, the required quotient = $\bm{3x^3 - \frac{8}{3}x^2 - 4 + \frac{1}{x}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{14p^2q^3 - 32p^3q^2 + 15pq^2 - 22p + 18q}{-2p^2q}\\[1em] \Rightarrow \frac{14p^2q^3}{-2p^2q} - \frac{32p^3q^2}{-2p^2q} + \frac{15pq^2}{-2p^2q} - \frac{22p}{-2p^2q} + \frac{18q}{-2p^2q}\\[1em] \Rightarrow -7q^{3-1} + 16p^{3-2}q^{2-1} - \frac{15q^{2-1}}{2p^{2-1}} + \frac{11}{pq} - \frac{9}{p^2}\\[1em] \Rightarrow -7q^2 + 16pq - \frac{15q}{2p} + \frac{11}{pq} - \frac{9}{p^2}\end{aligned}$$

Hence, the required quotient = $\bm{-7q^2 + 16pq - \frac{15q}{2p} + \frac{11}{pq} - \frac{9}{p^2}}$.

Dividing $6x^2 + 13x + 5$ by $2x + 1$

$$\begin{array}{l} \phantom{2x + 1\big)}{3x + 5} \\ 2x + 1\overline{\smash{\big)}6x^2 + 13x + 5} \\ \phantom{+++}\underline{\underset{-}{}6x^2 \underset{-}{+} 3x \phantom{+ 5}} \\ \phantom{++++++}10x + 5 \\ \phantom{+++++()}\underline{\underset{-}{}10x \underset{-}{+} 5} \\ \phantom{2x + 1\big)6x^2 + 13x;}0 \end{array}$$

Hence, quotient = $\bm{3x + 5}$ and remainder = 0.

Arranging the terms of the dividend and the divisor in descending powers of $y$, we divide $y^3 + 1$ by $y + 1$

$$\begin{array}{l} \phantom{y + 1\big)}{y^2 - y + 1} \\ y + 1\overline{\smash{\big)}y^3 + 1} \\ \phantom{++}\underline{\underset{-}{}y^3 + \underset{-}{}y^2 \phantom{+ 1}} \\ \phantom{y + 1\big)y^3;}-y^2 + 1 \\ \phantom{y + 1\big)y^3;}\underline{\underset{+}{-}y^2 ; -\underset{+}{}y \phantom{+ 1}} \\ \phantom{y + 1\big)y^3 + 1;}y + 1 \\ \phantom{y + 1\big)y^3 + 1;}\underline{\underset{-}{}y \underset{-}{+} 1} \\ \phantom{+++++++}0 \end{array}$$

Hence, quotient = $\bm{y^2 - y + 1}$ and remainder = 0.

Arranging the terms of the dividend in descending powers of $x$, we divide $-2x^2 + x + 5$ by $x + 1$

$$\begin{array}{l} \phantom{x + 1\big)}{-2x + 3} \\ x + 1\overline{\smash{\big)}-2x^2 + x + 5} \\ \phantom{x + 1\big)}\underline{\underset{+}{-}2x^2 \underset{+}{-} 2x \phantom{+ 5}} \\ \phantom{++++++}3x + 5 \\ \phantom{x + 1\big)-2x^2;}\underline{\underset{-}{}3x \underset{-}{+} 3} \\ \phantom{x + 1\big)-2x^2 + x;}2 \end{array}$$

Hence, quotient = $\bm{-2x + 3}$ and remainder = 2.

Dividing $x^3 - 6x^2 + 12x - 8$ by $x - 2$

$$\begin{array}{l} \phantom{x - 2\big)}{x^2 - 4x + 4} \\ x - 2\overline{\smash{\big)}x^3 - 6x^2 + 12x - 8} \\ \phantom{++}\underline{\underset{-}{}x^3 \underset{+}{-} 2x^2 \phantom{+ 12x} \phantom{- 8}} \\ \phantom{++++}-4x^2 + 12x - 8 \\ \phantom{x - 2\big)x^3;}\underline{\underset{+}{-}4x^2 \underset{-}{+} 8x \phantom{- 8}} \\ \phantom{++++++++}4x - 8 \\ \phantom{+++++++()}\underline{\underset{-}{}4x \underset{+}{-} 8} \\ \phantom{x - 2\big)x^3 - 6x^2 + 12x;}0 \end{array}$$

Hence, quotient = $\bm{x^2 - 4x + 4}$ and remainder = 0.

Dividing $6x^3 + x^2 - 26x - 25$ by $3x - 7$

$$\begin{array}{l} \phantom{3x - 7\big)}{2x^2 + 5x + 3} \\ 3x - 7\overline{\smash{\big)}6x^3 + x^2 - 26x - 25} \\ \phantom{+++}\underline{\underset{-}{}6x^3 \underset{+}{-} 14x^2 \phantom{- 26x} \phantom{- 25}} \\ \phantom{++++++}15x^2 - 26x - 25 \\ \phantom{+++++()}\underline{\underset{-}{}15x^2 \underset{+}{-} 35x \phantom{- 25}} \\ \phantom{++++++++++}9x - 25 \\ \phantom{+++++++++()}\underline{\underset{-}{}9x \underset{+}{-} 21} \\ \phantom{++++++++++++}-4 \end{array}$$

Hence, quotient = $\bm{2x^2 + 5x + 3}$ and remainder = -4.

Dividing $m^3 - 6m^2 + 7$ by $m - 1$

$$\begin{array}{l} \phantom{m - 1\big)}{m^2 - 5m - 5} \\ m - 1\overline{\smash{\big)}m^3 - 6m^2 + 7} \\ \phantom{++()}\underline{\underset{-}{}m^3 \underset{+}{-} m^2 \phantom{+ 7}} \\ \phantom{m - 1\big)m^3;}-5m^2 + 7 \\ \phantom{+++++}\underline{\underset{+}{-}5m^2 + \underset{-}{}5m \phantom{+ 7}} \\ \phantom{m - 1\big)m^3 - 6m^2;}-5m + 7 \\ \phantom{++++++++}\underline{\underset{+}{-}5m \underset{-}{+} 5} \\ \phantom{+++++++++++}2 \end{array}$$

Hence, quotient = $\bm{m^2 - 5m - 5}$ and remainder = 2.

Dividing $a^3 + 2a^2 + 2a + 1$ by $a^2 + a + 1$

$$\begin{array}{l} \phantom{a^2 + a + 1\big)}{a + 1} \\ a^2 + a + 1\overline{\smash{\big)}a^3 + 2a^2 + 2a + 1} \\ \phantom{a^2 + a + 1\big)}\underline{\underset{-}{}a^3 \underset{-}{+} a^2 \underset{-}{+} a \phantom{+ 1}} \\ \phantom{++++++++}a^2 + a + 1 \\ \phantom{++++++++}\underline{\underset{-}{}a^2 \underset{-}{+} a \underset{-}{+} 1} \\ \phantom{a^2 + a + 1\big)a^3 + 2a^2 + 2a;}0 \end{array}$$

Hence, quotient = $\bm{a + 1}$ and remainder = 0.

Dividing $12x^3 - 17x^2 + 26x - 18$ by $3x^2 - 2x + 5$

$$\begin{array}{l} \phantom{3x^2 - 2x + 5\big)}{4x - 3} \\ 3x^2 - 2x + 5\overline{\smash{\big)}12x^3 - 17x^2 + 26x - 18} \\ \phantom{3x^2 - 2x + 5\big)}\underline{\underset{-}{}12x^3 \underset{+}{-} 8x^2 \underset{-}{+} 20x \phantom{- 18}} \\ \phantom{+++++++++}-9x^2 + 6x - 18 \\ \phantom{+++++++++()}\underline{\underset{+}{-}9x^2 \underset{-}{+} 6x \underset{+}{-} 15} \\ \phantom{3x^2 - 2x + 5\big)12x^3 - 17x^2 + 26x;}-3 \end{array}$$

Hence, quotient = $\bm{4x - 3}$ and remainder = -3.

Area of the rectangle = $8x^2 - 45y^2 + 18xy$

One side = $4x + 15y$

As we know, Area of rectangle = One side × Adjacent side

∴ Adjacent side = Area of rectangle ÷ One side

Arranging the terms of the dividend in descending powers of $x$, we divide $8x^2 + 18xy - 45y^2$ by $4x + 15y$

$$\begin{array}{l} \phantom{4x + 15y\big)}{2x - 3y} \\ 4x + 15y\overline{\smash{\big)}8x^2 + 18xy - 45y^2} \\ \phantom{4x + 15y\big)}\underline{\underset{-}{}8x^2 \underset{-}{+} 30xy \phantom{- 45y^2}} \\ \phantom{4x + 15y\big)8x^2;}-12xy - 45y^2 \\ \phantom{4x + 15y\big)8x^2;}\underline{\underset{+}{-}12xy \underset{+}{-} 45y^2} \\ \phantom{4x + 15y\big)8x^2 + 18xy;}0 \end{array}$$

Hence, the length of the adjacent side = $\bm{2x - 3y}$.

If a polynomial is divided by another polynomial, then on subtracting the remainder from the dividend, the resulting polynomial becomes exactly divisible by the divisor.

So, we first find the remainder on dividing $x^4 + 2x^3 + x^2 - 18x - 15$ by $x^2 - 1 - x$.

Arranging the terms of the divisor in descending powers of $x$, we divide $x^4 + 2x^3 + x^2 - 18x - 15$ by $x^2 - x - 1$

$$\begin{array}{l} \phantom{x^2 - x - 1\big)}{x^2 + 3x + 5} \\ x^2 - x - 1\overline{\smash{\big)}x^4 + 2x^3 + x^2 - 18x - 15} \\ \phantom{x^2 - x - 1\big)}\underline{\underset{-}{}x^4 \underset{+}{-} x^3 \underset{+}{-} x^2 \phantom{- 18x} \phantom{- 15}} \\ \phantom{+++++++)}3x^3 + 2x^2 - 18x - 15 \\ \phantom{+++++++}\underline{\underset{-}{}3x^3 \underset{+}{-} 3x^2 \underset{+}{-} 3x \phantom{- 15}} \\ \phantom{++++++++++}5x^2 - 15x - 15 \\ \phantom{+++++++++()}\underline{\underset{-}{}5x^2 \underset{+}{-} 5x \underset{+}{-} 5} \\ \phantom{++++++++++++}-10x - 10 \end{array}$$

Here, quotient = $x^2 + 3x + 5$ and remainder = $-10x - 10$

Hence, $\bm{-10x - 10}$ should be subtracted from the given polynomial.

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$,

$$\begin{aligned}\Rightarrow (3x + 5)(3x + 5)\\[1em] \Rightarrow (3x + 5)^2\\[1em] \Rightarrow (3x)^2 + 2 \times 3x \times 5 + (5)^2\\[1em] \Rightarrow 9x^2 + 30x + 25\end{aligned}$$

Hence, $\bm{(3x + 5)(3x + 5) = 9x^2 + 30x + 25}$.

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$$\begin{aligned}\Rightarrow (9y - 5)(9y - 5)\\[1em] \Rightarrow (9y - 5)^2\\[1em] \Rightarrow (9y)^2 - 2 \times 9y \times 5 + (5)^2\\[1em] \Rightarrow 81y^2 - 90y + 25\end{aligned}$$

Hence, $\bm{(9y - 5)(9y - 5) = 81y^2 - 90y + 25}$.

Using the identity $(a + b)(a - b) = a^2 - b^2$,

$$\begin{aligned}\Rightarrow (4x + 11y)(4x - 11y)\\[1em] \Rightarrow (4x)^2 - (11y)^2\\[1em] \Rightarrow 16x^2 - 121y^2\end{aligned}$$

Hence, $\bm{(4x + 11y)(4x - 11y) = 16x^2 - 121y^2}$.

Using the identity $(a + b)(a - b) = a^2 - b^2$,

$$\begin{aligned}\Rightarrow \Big(\frac{3}{2}m + \frac{2}{3}n\Big)\Big(\frac{3}{2}m - \frac{2}{3}n\Big)\\[1em] \Rightarrow \Big(\frac{3}{2}m\Big)^2 - \Big(\frac{2}{3}n\Big)^2\\[1em] \Rightarrow \frac{9}{4}m^2 - \frac{4}{9}n^2\end{aligned}$$

Hence, $\bm{\Big(\frac{3}{2}m + \frac{2}{3}n\Big)\Big(\frac{3}{2}m - \frac{2}{3}n\Big) = \frac{9}{4}m^2 - \frac{4}{9}n^2}$.

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$,

$$\begin{aligned}\Rightarrow \Big(\frac{2}{a} + \frac{5}{b}\Big)\Big(\frac{2}{a} + \frac{5}{b}\Big)\\[1em] \Rightarrow \Big(\frac{2}{a} + \frac{5}{b}\Big)^2\\[1em] \Rightarrow \Big(\frac{2}{a}\Big)^2 + 2 \times \frac{2}{a} \times \frac{5}{b} + \Big(\frac{5}{b}\Big)^2\\[1em] \Rightarrow \frac{4}{a^2} + \frac{20}{ab} + \frac{25}{b^2}\end{aligned}$$

Hence, $\bm{\Big(\frac{2}{a} + \frac{5}{b}\Big)\Big(\frac{2}{a} + \frac{5}{b}\Big) = \frac{4}{a^2} + \frac{20}{ab} + \frac{25}{b^2}}$.

Using the identity $(a + b)(a - b) = a^2 - b^2$,

$$\begin{aligned}\Rightarrow \Big(\frac{p^2}{2} + \frac{2}{q^2}\Big)\Big(\frac{p^2}{2} - \frac{2}{q^2}\Big)\\[1em] \Rightarrow \Big(\frac{p^2}{2}\Big)^2 - \Big(\frac{2}{q^2}\Big)^2\\[1em] \Rightarrow \frac{p^4}{4} - \frac{4}{q^4}\end{aligned}$$

Hence, $\bm{\Big(\frac{p^2}{2} + \frac{2}{q^2}\Big)\Big(\frac{p^2}{2} - \frac{2}{q^2}\Big) = \frac{p^4}{4} - \frac{4}{q^4}}$.

⇒ 81²

⇒ (80 + 1)²

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$

⇒ (80)² + 2 × 80 × 1 + (1)²

⇒ 6400 + 160 + 1

⇒ 6561

Hence, 81² = 6561.

⇒ 97²

⇒ (100 - 3)²

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$

⇒ (100)² - 2 × 100 × 3 + (3)²

⇒ 10000 - 600 + 9

⇒ 9409

Hence, 97² = 9409.

⇒ 105²

⇒ (100 + 5)²

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$

⇒ (100)² + 2 × 100 × 5 + (5)²

⇒ 10000 + 1000 + 25

⇒ 11025

Hence, 105² = 11025.

⇒ 997²

⇒ (1000 - 3)²

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$

⇒ (1000)² - 2 × 1000 × 3 + (3)²

⇒ 1000000 - 6000 + 9

⇒ 994009

Hence, 997² = 994009.

⇒ 6.1²

⇒ (6 + 0.1)²

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$

⇒ (6)² + 2 × 6 × 0.1 + (0.1)²

⇒ 36 + 1.2 + 0.01

⇒ 37.21

Hence, 6.1² = 37.21.

⇒ 496 × 504

⇒ (500 - 4)(500 + 4)

Using the identity $(a - b)(a + b) = a^2 - b^2$

⇒ (500)² - (4)²

⇒ 250000 - 16

⇒ 249984

Hence, 496 × 504 = 249984.

⇒ 20.5 × 19.5

⇒ (20 + 0.5)(20 - 0.5)

Using the identity $(a + b)(a - b) = a^2 - b^2$

⇒ (20)² - (0.5)²

⇒ 400 - 0.25

⇒ 399.75

Hence, 20.5 × 19.5 = 399.75.

⇒ 9.6²

⇒ (10 - 0.4)²

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$

⇒ (10)² - 2 × 10 × 0.4 + (0.4)²

⇒ 100 - 8 + 0.16

⇒ 92.16

Hence, 9.6² = 92.16.

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$,

$$\begin{aligned}\Rightarrow (pq + 5r)^2\\[1em] \Rightarrow (pq)^2 + 2 \times pq \times 5r + (5r)^2\\[1em] \Rightarrow p^2q^2 + 10pqr + 25r^2\end{aligned}$$

Hence, $\bm{(pq + 5r)^2 = p^2q^2 + 10pqr + 25r^2}$.

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$$\begin{aligned}\Rightarrow \Big(\frac{5}{2}a - \frac{3}{5}b\Big)^2\\[1em] \Rightarrow \Big(\frac{5}{2}a\Big)^2 - 2 \times \frac{5}{2}a \times \frac{3}{5}b + \Big(\frac{3}{5}b\Big)^2\\[1em] \Rightarrow \frac{25}{4}a^2 - 3ab + \frac{9}{25}b^2\end{aligned}$$

Hence, $\bm{\Big(\frac{5}{2}a - \frac{3}{5}b\Big)^2 = \frac{25}{4}a^2 - 3ab + \frac{9}{25}b^2}$.

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$,

$$\begin{aligned}\Rightarrow (\sqrt{2}a + \sqrt{3}b)^2\\[1em] \Rightarrow (\sqrt{2}a)^2 + 2 \times \sqrt{2}a \times \sqrt{3}b + (\sqrt{3}b)^2\\[1em] \Rightarrow 2a^2 + 2\sqrt{6}ab + 3b^2\end{aligned}$$

Hence, $\bm{(\sqrt{2}a + \sqrt{3}b)^2 = 2a^2 + 2\sqrt{6}ab + 3b^2}$.

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$$\begin{aligned}\Rightarrow \Big(\frac{2x}{3y} - \frac{3y}{2x}\Big)^2\\[1em] \Rightarrow \Big(\frac{2x}{3y}\Big)^2 - 2 \times \frac{2x}{3y} \times \frac{3y}{2x} + \Big(\frac{3y}{2x}\Big)^2\\[1em] \Rightarrow \frac{4x^2}{9y^2} - 2 + \frac{9y^2}{4x^2}\end{aligned}$$

Hence, $\bm{\Big(\frac{2x}{3y} - \frac{3y}{2x}\Big)^2 = \frac{4x^2}{9y^2} - 2 + \frac{9y^2}{4x^2}}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (x + 7)(x + 3)\\[1em] \Rightarrow x^2 + (7 + 3)x + 7 \times 3\\[1em] \Rightarrow x^2 + 10x + 21\end{aligned}$$

Hence, $\bm{(x + 7)(x + 3) = x^2 + 10x + 21}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (3x + 4)(3x - 5)\\[1em] \Rightarrow (3x + 4)(3x + (-5))\\[1em] \Rightarrow (3x)^2 + (4 + (-5))(3x) + 4 \times (-5)\\[1em] \Rightarrow 9x^2 - 3x - 20\end{aligned}$$

Hence, $\bm{(3x + 4)(3x - 5) = 9x^2 - 3x - 20}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (p^2 + 2q)(p^2 - 3q)\\[1em] \Rightarrow (p^2 + 2q)(p^2 + (-3q))\\[1em] \Rightarrow (p^2)^2 + (2q + (-3q))p^2 + 2q \times (-3q)\\[1em] \Rightarrow p^4 - p^2q - 6q^2\end{aligned}$$

Hence, $\bm{(p^2 + 2q)(p^2 - 3q) = p^4 - p^2q - 6q^2}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (abc + 3)(abc - 5)\\[1em] \Rightarrow (abc + 3)(abc + (-5))\\[1em] \Rightarrow (abc)^2 + (3 + (-5))abc + 3 \times (-5)\\[1em] \Rightarrow a^2b^2c^2 - 2abc - 15\end{aligned}$$

Hence, $\bm{(abc + 3)(abc - 5) = a^2b^2c^2 - 2abc - 15}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow 203 \times 204\\[1em] \Rightarrow (200 + 3)(200 + 4)\\[1em] \Rightarrow (200)^2 + (3 + 4) \times 200 + 3 \times 4\\[1em] \Rightarrow 40000 + 1400 + 12\\[1em] \Rightarrow 41412\end{aligned}$$

Hence, 203 × 204 = 41412.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow 8.2 \times 8.7\\[1em] \Rightarrow (8 + 0.2)(8 + 0.7)\\[1em] \Rightarrow (8)^2 + (0.2 + 0.7) \times 8 + 0.2 \times 0.7\\[1em] \Rightarrow 64 + 7.2 + 0.14\\[1em] \Rightarrow 71.34\end{aligned}$$

Hence, 8.2 × 8.7 = 71.34.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow 107 \times 93\\[1em] \Rightarrow (100 + 7)(100 + (-7))\\[1em] \Rightarrow (100)^2 + (7 + (-7)) \times 100 + 7 \times (-7)\\[1em] \Rightarrow 10000 + 0 - 49\\[1em] \Rightarrow 9951\end{aligned}$$

Hence, 107 × 93 = 9951.

⇒ 53² - 47²

Using the identity $a^2 - b^2 = (a + b)(a - b)$

⇒ (53 + 47)(53 - 47)

⇒ 100 × 6

⇒ 600

Hence, 53² - 47² = 600.

⇒ (2.05)² - (0.95)²

Using the identity $a^2 - b^2 = (a + b)(a - b)$

⇒ (2.05 + 0.95)(2.05 - 0.95)

⇒ 3 × 1.1

⇒ 3.3

Hence, (2.05)² - (0.95)² = 3.3.

⇒ (14.3)² - (5.7)²

Using the identity $a^2 - b^2 = (a + b)(a - b)$

⇒ (14.3 + 5.7)(14.3 - 5.7)

⇒ 20 × 8.6

⇒ 172

Hence, (14.3)² - (5.7)² = 172.

Using the identities $(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$,

$$\begin{aligned}\Rightarrow (2x + 5y)^2 + (2x - 5y)^2\\[1em] \Rightarrow [(2x)^2 + 2 \times 2x \times 5y + (5y)^2] + [(2x)^2 - 2 \times 2x \times 5y + (5y)^2]\\[1em] \Rightarrow 4x^2 + 20xy + 25y^2 + 4x^2 - 20xy + 25y^2\\[1em] \Rightarrow (4x^2 + 4x^2) + (20xy - 20xy) + (25y^2 + 25y^2)\\[1em] \Rightarrow 8x^2 + 50y^2\end{aligned}$$

Hence, $\bm{(2x + 5y)^2 + (2x - 5y)^2 = 8x^2 + 50y^2}$.

Using the identity $a^2 - b^2 = (a + b)(a - b)$,

$$\begin{aligned}\Rightarrow \Big(\frac{7}{2}a - \frac{5}{2}b\Big)^2 - \Big(\frac{5}{2}a - \frac{7}{2}b\Big)^2\\[1em] \Rightarrow \Big[\Big(\frac{7}{2}a - \frac{5}{2}b\Big) + \Big(\frac{5}{2}a - \frac{7}{2}b\Big)\Big]\Big[\Big(\frac{7}{2}a - \frac{5}{2}b\Big) - \Big(\frac{5}{2}a - \frac{7}{2}b\Big)\Big]\\[1em] \Rightarrow \Big[\frac{7}{2}a + \frac{5}{2}a - \frac{5}{2}b - \frac{7}{2}b\Big]\Big[\frac{7}{2}a - \frac{5}{2}a - \frac{5}{2}b + \frac{7}{2}b\Big]\\[1em] \Rightarrow [6a - 6b][a + b]\\[1em] \Rightarrow 6(a - b)(a + b)\\[1em] \Rightarrow 6(a^2 - b^2)\\[1em] \Rightarrow 6a^2 - 6b^2\end{aligned}$$

Hence, $\bm{\Big(\frac{7}{2}a - \frac{5}{2}b\Big)^2 - \Big(\frac{5}{2}a - \frac{7}{2}b\Big)^2 = 6a^2 - 6b^2}$.

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$$\begin{aligned}\Rightarrow (p^2 - q^2r)^2 + 2p^2q^2r\\[1em] \Rightarrow [(p^2)^2 - 2 \times p^2 \times q^2r + (q^2r)^2] + 2p^2q^2r\\[1em] \Rightarrow p^4 - 2p^2q^2r + q^4r^2 + 2p^2q^2r\\[1em] \Rightarrow p^4 + q^4r^2\end{aligned}$$

Hence, $\bm{(p^2 - q^2r)^2 + 2p^2q^2r = p^4 + q^4r^2}$.

Taking L.H.S. and using the identity $a^2 - b^2 = (a + b)(a - b)$,

$$\begin{aligned}\Rightarrow (4x + 7y)^2 - (4x - 7y)^2\\[1em] \Rightarrow [(4x + 7y) + (4x - 7y)][(4x + 7y) - (4x - 7y)]\\[1em] \Rightarrow [4x + 7y + 4x - 7y][4x + 7y - 4x + 7y]\\[1em] \Rightarrow [8x][14y]\\[1em] \Rightarrow 112xy\\[1em] \Rightarrow \text{R.H.S.}\end{aligned}$$

Hence, proved that $\bm{(4x + 7y)^2 - (4x - 7y)^2 = 112xy}$.

Taking L.H.S. and using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$$\begin{aligned}\Rightarrow \Big(\frac{3}{7}p - \frac{7}{6}q\Big)^2 + pq\\[1em] \Rightarrow \Big(\frac{3}{7}p\Big)^2 - 2 \times \frac{3}{7}p \times \frac{7}{6}q + \Big(\frac{7}{6}q\Big)^2 + pq\\[1em] \Rightarrow \frac{9}{49}p^2 - \frac{2 \times 3 \times 7}{7 \times 6}pq + \frac{49}{36}q^2 + pq\\[1em] \Rightarrow \frac{9}{49}p^2 - pq + \frac{49}{36}q^2 + pq\\[1em] \Rightarrow \frac{9}{49}p^2 + \frac{49}{36}q^2\\[1em] \Rightarrow \text{R.H.S.}\end{aligned}$$

Hence, proved that $\bm{\Big(\frac{3}{7}p - \frac{7}{6}q\Big)^2 + pq = \frac{9}{49}p^2 + \frac{49}{36}q^2}$.

Taking L.H.S. and using the identity $(a + b)(a - b) = a^2 - b^2$,

$$\begin{aligned}\Rightarrow (p - q)(p + q) + (q - r)(q + r) + (r - p)(r + p)\\[1em] \Rightarrow (p^2 - q^2) + (q^2 - r^2) + (r^2 - p^2)\\[1em] \Rightarrow p^2 - q^2 + q^2 - r^2 + r^2 - p^2\\[1em] \Rightarrow (p^2 - p^2) + (-q^2 + q^2) + (-r^2 + r^2)\\[1em] \Rightarrow 0\\[1em] \Rightarrow \text{R.H.S.}\end{aligned}$$

Hence, proved that $\bm{(p - q)(p + q) + (q - r)(q + r) + (r - p)( r + p) = 0}$.

(i) Given, $x - \frac{1}{x} = 7$

Squaring both sides,

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 7^2\\[1em] \Rightarrow x^2 - 2 \times x \times \frac{1}{x} + \frac{1}{x^2} = 49\\[1em] \Rightarrow x^2 - 2 + \frac{1}{x^2} = 49\\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 49 + 2\\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 51\end{aligned}$$

Hence, $\bm{x^2 + \frac{1}{x^2} = 51}$.

(ii) From part (i),

$x^2 + \frac{1}{x^2} = 51$

Squaring both sides,

$$\begin{aligned}\Rightarrow \Big(x^2 + \frac{1}{x^2}\Big)^2 = 51^2\\[1em] \Rightarrow (x^2)^2 + 2 \times x^2 \times \frac{1}{x^2} + \Big(\frac{1}{x^2}\Big)^2 = 2601\\[1em] \Rightarrow x^4 + 2 + \frac{1}{x^4} = 2601\\[1em] \Rightarrow x^4 + \frac{1}{x^4} = 2601 - 2\\[1em] \Rightarrow x^4 + \frac{1}{x^4} = 2599\end{aligned}$$

Hence, $\bm{x^4 + \frac{1}{x^4} = 2599}$.

(i) Given, $x^2 + \frac{1}{x^2} = 23$

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 = x^2 + 2 \times x \times \frac{1}{x} + \frac{1}{x^2}\\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big)^2 = x^2 + \frac{1}{x^2} + 2\\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big)^2 = 23 + 2\\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big)^2 = 25\\[1em] \Rightarrow x + \frac{1}{x} = \pm\sqrt{25}\\[1em] \Rightarrow x + \frac{1}{x} = \pm 5\end{aligned}$$

Hence, $\bm{x + \frac{1}{x} = \pm5}$.

(ii) Using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x}\Big)^2 = x^2 - 2 \times x \times \frac{1}{x} + \frac{1}{x^2}\\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = x^2 + \frac{1}{x^2} - 2\\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 23 - 2\\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 21\\[1em] \Rightarrow x - \frac{1}{x} = \pm\sqrt{21}\end{aligned}$$

Hence, $\bm{x - \frac{1}{x} = \pm \sqrt{21}}$.

Given, $a + b = 9$ and $ab = 10$

We know that $(a + b)^2 = a^2 + 2ab + b^2 = a^2 + b^2 + 2ab$

Substituting the values,

$$\begin{aligned}\Rightarrow (9)^2 = a^2 + b^2 + 2 \times 10\\[1em] \Rightarrow 81 = a^2 + b^2 + 20\\[1em] \Rightarrow a^2 + b^2 = 81 - 20\\[1em] \Rightarrow a^2 + b^2 = 61\end{aligned}$$

Hence, $\bm{a^2 + b^2 = 61}$.

Given, $a - b = 6$ and $a^2 + b^2 = 42$

We know that $(a - b)^2 = a^2 - 2ab + b^2 = a^2 + b^2 - 2ab$

Substituting the values,

$$\begin{aligned}\Rightarrow (6)^2 = 42 - 2ab\\[1em] \Rightarrow 36 = 42 - 2ab\\[1em] \Rightarrow 2ab = 42 - 36\\[1em] \Rightarrow 2ab = 6\\[1em] \Rightarrow ab = 3\end{aligned}$$

Hence, $\bm{ab = 3}$.

Given, $a^2 + b^2 = 41$ and $ab = 4$

(i) We know that $(a + b)^2 = a^2 + b^2 + 2ab$

Substituting the values,

$$\begin{aligned}\Rightarrow (a + b)^2 = 41 + 2 \times 4\\[1em] \Rightarrow (a + b)^2 = 41 + 8\\[1em] \Rightarrow (a + b)^2 = 49\\[1em] \Rightarrow a + b = \pm\sqrt{49}\\[1em] \Rightarrow a + b = \pm 7\end{aligned}$$

Hence, $\bm{a + b = \pm 7}$.

(ii) We know that $(a - b)^2 = a^2 + b^2 - 2ab$

Substituting the values,

$$\begin{aligned}\Rightarrow (a - b)^2 = 41 - 2 \times 4\\[1em] \Rightarrow (a - b)^2 = 41 - 8\\[1em] \Rightarrow (a - b)^2 = 33\\[1em] \Rightarrow a - b = \pm\sqrt{33}\end{aligned}$$

Hence, $\bm{a - b = \pm\sqrt{33}}$.

(i) A symbol which has a fixed value is called a constant.

(ii) A symbol which can be given various numerical values is called variable or literal.

(iii) The various parts of an algebraic expression separated by + or - sign are called terms.

(iv) An algebraic expression having two terms is called a binomial.

(v) Each of the quantity (constant or literal) multiplied together to form a product is called a factor of the product.

(vi) The terms having same literal coefficients are called like terms otherwise they are called unlike terms.

(vii) Degree of the polynomial is the greatest sum of the powers of variables in each term.

(viii) An identity is an equality which is true for all values of variables in it.

(ix) $(a + b)^2 = \bm{a^2 + 2ab + b^2}$.

(x) $(x + a)(x + b) = x^2 + \bm{(a + b)x} + ab$.

(xi) Dividend = divisor × quotient + remainder.

(i) An algebraic expression having only one term is called a monomial.

True.

(ii) A symbol which has fixed value is called a literal.

False.

Reason — A symbol which has a fixed value is called a constant. A literal is a symbol which can be given various numerical values.

(iii) The term of an algebraic expression having no literal factor is called its constant term.

True.

(iv) In any term of an algebraic expression the constant part is called literal coefficient of the term.

False.

Reason — In any term the constant part is called the numerical coefficient of the term, while the remaining part is called the literal coefficient.

(v) An algebraic expression is called polynomial if the powers of the variables involved in it in each term are non-negative integers.

True.

(vi) $5x^2y^2z, 3x^2zy^2, -\frac{4}{5}zx^2y^2$ are unlike terms.

False.

Reason — The literal coefficient of each of the three terms is $x^2y^2z$. As the terms have the same literal coefficients, they are like terms.

(vii) $5x + \frac{2}{x} + 7$ is a polynomial of degree 1

False.

Reason — Here $\frac{2}{x} = 2x^{-1}$, in which the power of $x$ is -1, a negative integer. So, the expression is not a polynomial at all.

(viii) $3x^2y + 7x + 8y + 9$ is a polynomial of degree 3

True.

(ix) Numerical coefficient of $-7x^3y$ is 7

False.

Reason — The numerical coefficient of $-7x^3y$ is -7.

(x) $xyz + yzx + zxy$ is a monomial.

True.

(xi) An equation is true for all values of variables in it.

False.

Reason — An equation is true only for certain values of the variables in it. An identity is true for all values of the variables in it.

(xii) $(a - b)^2 + 2ab = a^2 + b^2$

True.

Any factor of a term is called the coefficient of the remaining factor of the term. The constant part is the numerical coefficient and the remaining part is the literal coefficient.

In $-9xyz^2$, the numerical coefficient is -9 and the remaining part is $xyz^2$.

Hence, option 3 is the correct option.

An algebraic expression is a polynomial if the powers of the variables in each term are non-negative integers.

Option 1: In $3x^2 - 2x + 7$ the powers of $x$ are 2, 1 and 0. So, it is a polynomial.

Option 2: $\frac{5x^3}{2x} + 3x^2 + 8 = \frac{5}{2}x^2 + 3x^2 + 8$, in which the powers of $x$ are 2 and 0. So, it is a polynomial.

Option 3: In $3x + \frac{2}{x} + 7$, the term $\frac{2}{x} = 2x^{-1}$, in which the power of $x$ is -1, a negative integer. So, it is not a polynomial.

Option 4: In $\sqrt{2}x^2 + \sqrt{3}x + \sqrt{6}$ the powers of $x$ are 2, 1 and 0. So, it is a polynomial.

Hence, option 3 is the correct option.

An algebraic expression having only one term is called a monomial. Multiplication and division do not separate the terms of an algebraic expression.

Option 1: $3x \times y \times z$ has only one term, so it is a monomial.

Option 2: $-5pq$ has only one term, so it is a monomial.

Option 3: $8m^2 \times n \div 3l$ has only one term, so it is a monomial.

Option 4: $7x + y - z$ has three terms, so it is not a monomial. It is a trinomial.

Hence, option 4 is the correct option.

In a polynomial in two or more variables, the degree is the greatest sum of the powers of the variables in a term.

The given polynomial is $7x^2yz^2 + 6x^3y^2z^2 - 5x + 8y$

Sum of powers of $7x^2yz^2 = 2 + 1 + 2 = 5$

Sum of powers of $6x^3y^2z^2 = 3 + 2 + 2 = 7$

Sum of powers of $-5x = 1$

Sum of powers of $8y = 1$

The greatest sum is 7.

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow a(b - c) + b(c - a) + c(a - b)\\[1em] \Rightarrow ab - ac + bc - ab + ac - bc\\[1em] \Rightarrow (ab - ab) + (-ac + ac) + (bc - bc)\\[1em] \Rightarrow 0\end{aligned}$$

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{7}{5}xy - \frac{2}{3}xy + \frac{8}{9}xy\\[1em] \Rightarrow \Big(\frac{7}{5} - \frac{2}{3} + \frac{8}{9}\Big)xy\\[1em] \Rightarrow \frac{63 - 30 + 40}{45}xy\\[1em] \Rightarrow \frac{73}{45}xy\end{aligned}$$

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (3p^2qr^3) \times (-4p^3q^2r^2) \times (7pq^2r)\\[1em] \Rightarrow (3 \times (-4) \times 7) \times (p^2 \times p^3 \times p) \times (q \times q^2 \times q^2) \times (r^3 \times r^2 \times r)\\[1em] \Rightarrow -84p^{(2+3+1)}q^{(1+2+2)}r^{(3+2+1)}\\[1em] \Rightarrow -84p^6q^5r^6\end{aligned}$$

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 3m \times (2m^2 - 5mn + 4n^2)\\[1em] \Rightarrow 3m \times 2m^2 - 3m \times 5mn + 3m \times 4n^2\\[1em] \Rightarrow 6m^3 - 15m^2n + 12mn^2\end{aligned}$$

Hence, option 2 is the correct option.

Volume of a rectangular box = length × breadth × height

$$\begin{aligned}\Rightarrow 3x^2y \times 4y^2z \times 5z^2x\\[1em] \Rightarrow (3 \times 4 \times 5) \times (x^2 \times x) \times (y \times y^2) \times (z \times z^2)\\[1em] \Rightarrow 60x^3y^3z^3\end{aligned}$$

Hence, option 3 is the correct option.

$(x + 1)(x + 2)$

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow x^2 + (1 + 2)x + 1 \times 2\\[1em] \Rightarrow x^2 + 3x + 2\end{aligned}$$

Hence, option 3 is the correct option.

Given, $x + \frac{1}{x} = 2$

Squaring both sides,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 = 2^2\\[1em] \Rightarrow x^2 + 2 \times x \times \frac{1}{x} + \frac{1}{x^2} = 4\\[1em] \Rightarrow x^2 + 2 + \frac{1}{x^2} = 4\\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 4 - 2\\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 2\end{aligned}$$

Hence, option 2 is the correct option.

Given,

$x^2 + y^2 = 9$ and $xy = 8$

We know that,

$(x + y)^2 = x^2 + y^2 + 2xy$

Substituting the values,

$$\begin{aligned}\Rightarrow (x + y)^2 = 9 + 2 \times 8\\[1em] \Rightarrow (x + y)^2 = 9 + 16\\[1em] \Rightarrow (x + y)^2 = 25\\[1em] \Rightarrow x + y = \pm\sqrt{25}\\[1em] \Rightarrow x + y = \pm 5\end{aligned}$$

Hence, option 4 is the correct option.

Using the identity,

$a^2 - b^2 = (a + b)(a - b)$

⇒ (102)² - (98)²

⇒ (102 + 98)(102 - 98)

⇒ 200 × 4

⇒ 800

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow -50x^3y^2z^2 \div (-5xyz)\\[1em] \Rightarrow \frac{-50x^3y^2z^2}{-5xyz}\\[1em] \Rightarrow \frac{50}{5} \times x^{3-1}y^{2-1}z^{2-1}\\[1em] \Rightarrow 10x^2yz\end{aligned}$$

Hence, option 2 is the correct option.

$96 \times 104 = (100 - 4)(100 + 4)$

Using the identity $(a - b)(a + b) = a^2 - b^2$,

$$\begin{aligned}\Rightarrow (100)^2 - (4)^2\\[1em] \Rightarrow 10000 - 16\\[1em] \Rightarrow 9984\end{aligned}$$

Hence, option 1 is the correct option.

Area of rectangle = length × breadth

⇒ Breadth = Area of rectangle ÷ length

$$\begin{aligned}\Rightarrow \frac{24(x^2yz + xy^2z + xyz^2)}{8xyz}\\[1em] \Rightarrow \frac{24xyz(x + y + z)}{8xyz}\\[1em] \Rightarrow 3(x + y + z)\end{aligned}$$

Hence, option 1 is the correct option.

Volume of a cube = (edge)³

$$\begin{aligned}\Rightarrow (x + y)^3\\[1em] \Rightarrow (x + y)(x + y)^2\\[1em] \Rightarrow (x + y)(x^2 + 2xy + y^2)\\[1em] \Rightarrow x^3 + 2x^2y + xy^2 + x^2y + 2xy^2 + y^3\\[1em] \Rightarrow x^3 + y^3 + 3x^2y + 3xy^2\\[1em] \Rightarrow x^3 + y^3 + 3xy(x + y)\end{aligned}$$

So, Statement I is true.

Volume of a cube = (edge)³ = $a^3$, where $a$ is the edge of the cube.

So, Statement II is true.

Hence, option 3 is the correct option.

Area of the rectangle = $x^2 + y^2 + 2xy$

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$,

$$\begin{aligned}\Rightarrow (x + y)^2\\[1em] \Rightarrow (x + y)(x + y)\end{aligned}$$

As one side is $(x + y)$, the adjacent side = $(x + y)$.

As both the adjacent sides are equal, the rectangle is a square.

So, Statement I is true.

We know that $(x - y)^2 = x^2 + y^2 - 2xy$ and not $x^2 + y^2 + 2xy$.

So, Statement II is false.

Hence, option 1 is the correct option.

Dividing $2x^4 + 5x^3 - 18x^2 + 6x + 8$ by $x^2 + 5x + 2$

$$\begin{array}{l} \phantom{x^2 + 5x + 2\big)}{2x^2 - 5x + 3} \\ x^2 + 5x + 2\overline{\smash{\big)}2x^4 + 5x^3 - 18x^2 + 6x + 8} \\ \phantom{+++++}\underline{\underset{-}{}2x^4 \underset{-}{+} 10x^3 \underset{-}{+} 4x^2 \phantom{+ 6x} \phantom{+ 8}} \\ \phantom{+++++++}-5x^3 - 22x^2 + 6x + 8 \\ \phantom{x^2 + 5x + 2\big)2x^4;}\underline{\underset{+}{-}5x^3 \underset{+}{-} 25x^2 \underset{+}{-} 10x \phantom{+ 8}} \\ \phantom{+++++++++++}3x^2 + 16x + 8 \\ \phantom{x^2 + 5x + 2\big)2x^4 + 5x^3;}\underline{\underset{-}{}3x^2 \underset{-}{+} 15x \underset{-}{+} 6} \\ \phantom{++++++++++++++}x + 2 \end{array}$$

Here, quotient = $2x^2 - 5x + 3$ and remainder = $x + 2$

But Statement I says the quotient is $2x^2 + 5x + 3$.

So, Statement I is false.

We know that Dividend = Divisor × Quotient + Remainder

⇒ Dividend - Remainder = Divisor × Quotient

⇒ Divisor = $\frac{\text{Dividend} - \text{Remainder}}{\text{Quotient}}$

So, Statement II is true.

Hence, option 2 is the correct option.

Given, $x - \frac{1}{x} = 1$

Using the identity $a^2 - b^2 = (a + b)(a - b)$,

$$\begin{aligned}\Rightarrow x^2 - \frac{1}{x^2}\\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big)\Big(x - \frac{1}{x}\Big)\\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big) \times 1\\[1em] \Rightarrow x + \frac{1}{x}\end{aligned}$$

So, Statement I is true.

We know that $(a - b)^2 = a^2 + b^2 - 2ab$ and not $a^2 - b^2 - 2ab$.

So, Statement II is false.

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (x + 3)(x - 3)(x + 1)(x - 1)\\[1em] \Rightarrow [(x + 3)(x - 3)][(x + 1)(x - 1)]\end{aligned}$$

Using the identity $(a + b)(a - b) = a^2 - b^2$,

$$\begin{aligned}\Rightarrow (x^2 - 9)(x^2 - 1)\\[1em] \Rightarrow x^4 - x^2 - 9x^2 + 9\\[1em] \Rightarrow x^4 - 10x^2 + 9\end{aligned}$$

The greatest power of $x$ is 4, so it is a polynomial of degree 4 and not of degree 1.

So, Statement I is false.

A polynomial is of degree $n$ if the greatest sum of the powers of variables in the terms is $n$.

So, Statement II is true.

Hence, option 2 is the correct option.

Write each expression in a separate row, keeping the like terms one below the other, and then add them.

$$\begin{array}{rrrrr} & - 5x^2y & + 3xy^2 & - 7xy & + 8 \\ + & 12x^2y & - 5xy^2 & + 3xy & - 2 \\ \hline & 7x^2y & - 2xy^2 & - 4xy & + 6 \\ \end{array}$$

Hence, the required sum = $\bm{7x^2y - 2xy^2 - 4xy + 6}$.

As $zx$ is the same as $xz$, the third expression can be written as $-5zx + 2x - 5xy$.

Write each expression in a separate row, keeping the like terms one below the other, and then add them.

$$\begin{array}{rrrrrr} & 9xy & + 3yz & - 5zx & & \\ & & + 4yz & + 9zx & & - 5y \\ + & - 5xy & & - 5zx & + 2x & \\ \hline & 4xy & + 7yz & - zx & + 2x & - 5y \\ \end{array}$$

Hence, the required sum = $\bm{4xy + 7yz - zx + 2x - 5y}$.

Subtracting,

$$\begin{array}{rrrrr} & 3a & + 5b & - 9c & + 3 \\ & 5a & - 3b & + 11c & - 2 \\ &- &+ &- &+ \\ \hline & - 2a & + 8b & - 20c & + 5 \\ \end{array}$$

Hence, the required difference = $\bm{-2a + 8b - 20c + 5}$.

Subtracting,

$$\begin{array}{rrrrrrr} & 8x^2 & - 5xy & + 2y^2 & + 5x & - 3y & \\ & 10x^2 & & - 8y^2 & & + 5y & - 3 \\ &- & &+ & &- &+ \\ \hline & - 2x^2 & - 5xy & + 10y^2 & + 5x & - 8y & + 3 \\ \end{array}$$

Hence, the required difference = $\bm{-2x^2 - 5xy + 10y^2 + 5x - 8y + 3}$.

Let the required expression be A.

According to the question,

$(5x^2 - 3x + 1) + A = 3x^3 - 7x^2 + 8$

⇒ $A = (3x^3 - 7x^2 + 8) - (5x^2 - 3x + 1)$

Subtracting,

$$\begin{array}{rrrrr} & 3x^3 & - 7x^2 & & + 8 \\ & & 5x^2 & - 3x & + 1 \\ & &- &+ &- \\ \hline & 3x^3 & - 12x^2 & + 3x & + 7 \\ \end{array}$$

Hence, $\bm{3x^3 - 12x^2 + 3x + 7}$ must be added.

Solving,

$$\begin{aligned}\Rightarrow 3x^2y \times (-4xy^2)\\[1em] \Rightarrow (3 \times (-4)) \times (x^2 \times x) \times (y \times y^2)\\[1em] \Rightarrow -12x^{(2+1)}y^{(1+2)}\\[1em] \Rightarrow -12x^3y^3\end{aligned}$$

Hence, the required product = $\bm{-12x^3y^3}$.

Solving,

$$\begin{aligned}\Rightarrow -\frac{4}{5}xy \times \frac{5}{7}yz \times \Big(-\frac{14}{9}zx\Big)\\[1em] \Rightarrow \Big(-\frac{4}{5} \times \frac{5}{7} \times -\frac{14}{9}\Big) \times (x \times x) \times (y \times y) \times (z \times z)\\[1em] \Rightarrow \frac{4 \times 5 \times 14}{5 \times 7 \times 9}x^2y^2z^2\\[1em] \Rightarrow \frac{280}{315}x^2y^2z^2\\[1em] \Rightarrow \frac{8}{9}x^2y^2z^2\end{aligned}$$

Hence, the required product = $\bm{\frac{8}{9}x^2y^2z^2}$.

Solving,

$$\begin{aligned}\Rightarrow (3pq - 4p^2 + 5q^2 + 7) \times (-7pq)\\[1em] \Rightarrow 3pq \times (-7pq) - 4p^2 \times (-7pq) + 5q^2 \times (-7pq) + 7 \times (-7pq)\\[1em] \Rightarrow -21p^2q^2 + 28p^3q - 35pq^3 - 49pq\end{aligned}$$

Hence, the required product = $\bm{-21p^2q^2 + 28p^3q - 35pq^3 - 49pq}$.

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{3}{4}x^2y - \frac{4}{5}xy + \frac{5}{6}xy^2\Big) \times (-15xyz)\\[1em] \Rightarrow \frac{3}{4}x^2y \times (-15xyz) - \frac{4}{5}xy \times (-15xyz) + \frac{5}{6}xy^2 \times (-15xyz)\\[1em] \Rightarrow -\frac{3 \times 15}{4}x^3y^2z + \frac{4 \times 15}{5}x^2y^2z - \frac{5 \times 15}{6}x^2y^3z\\[1em] \Rightarrow -\frac{45}{4}x^3y^2z + 12x^2y^2z - \frac{25}{2}x^2y^3z\end{aligned}$$

Hence, the required product = $\bm{-\frac{45}{4}x^3y^2z + 12x^2y^2z - \frac{25}{2}x^2y^3z}$.

$(5x^2 + 4x - 2)(3 - x - 4x^2)$

Arranging the terms of the second polynomial in descending powers of $x$,

$$\begin{aligned}\Rightarrow (5x^2 + 4x - 2)(-4x^2 - x + 3)\\[1em] \Rightarrow 5x^2(-4x^2 - x + 3) + 4x(-4x^2 - x + 3) - 2(-4x^2 - x + 3)\\[1em] \Rightarrow -20x^4 - 5x^3 + 15x^2 - 16x^3 - 4x^2 + 12x + 8x^2 + 2x - 6\\[1em] \Rightarrow -20x^4 + (-5x^3 - 16x^3) + (15x^2 - 4x^2 + 8x^2) + (12x + 2x) - 6\\[1em] \Rightarrow -20x^4 - 21x^3 + 19x^2 + 14x - 6\end{aligned}$$

Hence, the required product = $\bm{-20x^4 - 21x^3 + 19x^2 + 14x - 6}$.

Solving,

$$\begin{aligned}\Rightarrow (7x^2 + 12xy - 9y^2)(3x^2 - 5xy + 3y^2)\\[1em] \Rightarrow 7x^2(3x^2 - 5xy + 3y^2) + 12xy(3x^2 - 5xy + 3y^2) - 9y^2(3x^2 - 5xy + 3y^2)\\[1em] \Rightarrow 21x^4 - 35x^3y + 21x^2y^2 + 36x^3y - 60x^2y^2 + 36xy^3 - 27x^2y^2 + 45xy^3 - 27y^4\\[1em] \Rightarrow 21x^4 + (-35x^3y + 36x^3y) + (21x^2y^2 - 60x^2y^2 - 27x^2y^2) + (36xy^3 + 45xy^3) - 27y^4\\[1em] \Rightarrow 21x^4 + x^3y - 66x^2y^2 + 81xy^3 - 27y^4\end{aligned}$$

Hence, the required product = $\bm{21x^4 + x^3y - 66x^2y^2 + 81xy^3 - 27y^4}$.

$(3ab - 2a^2 + 5b^2) \times (2b^2 - 5ab + 3a^2) + 8a^3b - 7b^4$

First we find the product of the two polynomials.

$$\begin{aligned}\Rightarrow (3ab - 2a^2 + 5b^2)(2b^2 - 5ab + 3a^2)\\[1em] \Rightarrow 3ab(2b^2 - 5ab + 3a^2) - 2a^2(2b^2 - 5ab + 3a^2) + 5b^2(2b^2 - 5ab + 3a^2)\\[1em] \Rightarrow 6ab^3 - 15a^2b^2 + 9a^3b - 4a^2b^2 + 10a^3b - 6a^4 + 10b^4 - 25ab^3 + 15a^2b^2\\[1em] \Rightarrow -6a^4 + (9a^3b + 10a^3b) + (-15a^2b^2 - 4a^2b^2 + 15a^2b^2) + (6ab^3 - 25ab^3) + 10b^4\\[1em] \Rightarrow -6a^4 + 19a^3b - 4a^2b^2 - 19ab^3 + 10b^4\end{aligned}$$

Now, adding $8a^3b - 7b^4$,

$$\begin{aligned}\Rightarrow -6a^4 + 19a^3b - 4a^2b^2 - 19ab^3 + 10b^4 + 8a^3b - 7b^4\\[1em] \Rightarrow -6a^4 + (19a^3b + 8a^3b) - 4a^2b^2 - 19ab^3 + (10b^4 - 7b^4)\\[1em] \Rightarrow 27a^3b - 4a^2b^2 - 19ab^3 - 6a^4 + 3b^4\end{aligned}$$

For $a = 1, b = -1$,

$$\begin{aligned}\Rightarrow 27(1)^3(-1) - 4(1)^2(-1)^2 - 19(1)(-1)^3 - 6(1)^4 + 3(-1)^4\\[1em] \Rightarrow -27 - 4 + 19 - 6 + 3\\[1em] \Rightarrow -15\end{aligned}$$

Hence, the simplified expression = $\bm{27a^3b - 4a^2b^2 - 19ab^3 - 6a^4 + 3b^4}$ and its value = -15.

Solving,

$$\begin{aligned}\Rightarrow (1.7x - 2.5y)(2y + 3x + 4) - 7.8x^2 - 10y\\[1em] \Rightarrow 1.7x(2y + 3x + 4) - 2.5y(2y + 3x + 4) - 7.8x^2 - 10y\\[1em] \Rightarrow 3.4xy + 5.1x^2 + 6.8x - 5y^2 - 7.5xy - 10y - 7.8x^2 - 10y\\[1em] \Rightarrow (5.1x^2 - 7.8x^2) + (3.4xy - 7.5xy) - 5y^2 + 6.8x + (-10y - 10y)\\[1em] \Rightarrow -2.7x^2 - 4.1xy - 5y^2 + 6.8x - 20y\end{aligned}$$

For $x = 0, y = 1$,

$$\begin{aligned}\Rightarrow -2.7(0)^2 - 4.1(0)(1) - 5(1)^2 + 6.8(0) - 20(1)\\[1em] \Rightarrow 0 - 0 - 5 + 0 - 20\\[1em] \Rightarrow -25\end{aligned}$$

Hence, the simplified expression = $\bm{-2.7x^2 - 4.1xy - 5y^2 + 6.8x - 20y}$ and its value = -25.

Solving,

$$\begin{aligned}\Rightarrow 66pq^2r^3 \div 11qr^2\\[1em] \Rightarrow \frac{66pq^2r^3}{11qr^2}\\[1em] \Rightarrow \frac{66}{11} \times p \times q^{2-1} \times r^{3-2}\\[1em] \Rightarrow 6pqr\end{aligned}$$

Hence, the required quotient = $\bm{6pqr}$.

Solving,

$$\begin{aligned}\Rightarrow (x^3 + 2x^2 + 3x) \div 2x\\[1em] \Rightarrow \frac{x^3}{2x} + \frac{2x^2}{2x} + \frac{3x}{2x}\\[1em] \Rightarrow \frac{x^{3-1}}{2} + x^{2-1} + \frac{3}{2}\\[1em] \Rightarrow \frac{x^2}{2} + x + \frac{3}{2}\end{aligned}$$

Hence, the required quotient = $\bm{\frac{x^2}{2} + x + \frac{3}{2}}$.

Dividing $10x^4 - 19x^3 + 17x^2 + 15x - 42$ by $2x^2 - 3x + 5$

$$\begin{array}{l} \phantom{2x^2 - 3x + 5\big)}{5x^2 - 2x - 7} \\ 2x^2 - 3x + 5\overline{\smash{\big)}10x^4 - 19x^3 + 17x^2 + 15x - 42} \\ \phantom{+++++()}\underline{\underset{-}{}10x^4 \underset{+}{-} 15x^3 \underset{-}{+} 25x^2 \phantom{+ 15x} \phantom{- 42}} \\ \phantom{+++++++++}-4x^3 - 8x^2 + 15x - 42 \\ \phantom{+++++++++()}\underline{\underset{+}{-}4x^3 \underset{-}{+} 6x^2 \underset{+}{-} 10x \phantom{- 42}} \\ \phantom{++++++++++++}-14x^2 + 25x - 42 \\ \phantom{++++++++++++}\underline{\underset{+}{-}14x^2 \underset{-}{+} 21x \underset{+}{-} 35} \\ \phantom{2x^2 - 3x + 5\big)10x^4 - 19x^3 + 17x^2 + 15x;}4x - 7 \end{array}$$

Hence, quotient = $\bm{5x^2 - 2x - 7}$ and remainder = $\bm{4x - 7}$.

$\Rightarrow (3x + 4y)(3x + 4y) = (3x + 4y)^2$

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$,

$$\begin{aligned}\Rightarrow (3x)^2 + 2 \times 3x \times 4y + (4y)^2\\[1em] \Rightarrow 9x^2 + 24xy + 16y^2\end{aligned}$$

Hence, $\bm{(3x + 4y)(3x + 4y) = 9x^2 + 24xy + 16y^2}$.

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$$\begin{aligned}\Rightarrow \Big(\frac{5}{2}a - b\Big)\Big(\frac{5}{2}a - b\Big)\\[1em] \Rightarrow \Big(\frac{5}{2}a - b\Big)^2\\[1em] \Rightarrow \Big(\frac{5}{2}a\Big)^2 - 2 \times \frac{5}{2}a \times b + (b)^2\\[1em] \Rightarrow \frac{25}{4}a^2 - 5ab + b^2\end{aligned}$$

Hence, $\bm{\Big(\frac{5}{2}a - b\Big)\Big(\frac{5}{2}a - b\Big) = \frac{25}{4}a^2 - 5ab + b^2}$.

$\Rightarrow (3.5m - 1.5n)(3.5m + 1.5n)$

Using the identity $(a - b)(a + b) = a^2 - b^2$,

$$\begin{aligned}\Rightarrow (3.5m)^2 - (1.5n)^2\\[1em] \Rightarrow 12.25m^2 - 2.25n^2\end{aligned}$$

Hence, $\bm{(3.5m - 1.5n)(3.5m + 1.5n) = 12.25m^2 - 2.25n^2}$.

$\Rightarrow (7xy - 2)(7xy + 7) = (7xy + (-2))(7xy + 7)$

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (7xy)^2 + ((-2) + 7)(7xy) + (-2) \times 7\\[1em] \Rightarrow 49x^2y^2 + 5 \times 7xy - 14\\[1em] \Rightarrow 49x^2y^2 + 35xy - 14\end{aligned}$$

Hence, $\bm{(7xy - 2)(7xy + 7) = 49x^2y^2 + 35xy - 14}$.

⇒ 105² = (100 + 5)²

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$,

⇒ (100)² + 2 × 100 × 5 + (5)²

⇒ 10000 + 1000 + 25

⇒ 11025

Hence, 105² = 11025.

⇒ 97² = (100 - 3)²

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

⇒ (100)² - 2 × 100 × 3 + (3)²

⇒ 10000 - 600 + 9

⇒ 9409

Hence, 97² = 9409.

⇒ 201 × 199 = (200 + 1)(200 - 1)

Using the identity $(a + b)(a - b) = a^2 - b^2$,

⇒ (200)² - (1)²

⇒ 40000 - 1

⇒ 39999

Hence, 201 × 199 = 39999.

⇒ 87² - 13²

Using the identity $a^2 - b^2 = (a + b)(a - b)$,

⇒ (87 + 13)(87 - 13)

⇒ 100 × 74

⇒ 7400

Hence, 87² - 13² = 7400.

⇒ 105 × 107 = (100 + 5)(100 + 7)

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

⇒ (100)² + (5 + 7) × 100 + 5 × 7

⇒ 10000 + 1200 + 35

⇒ 11235

Hence, 105 × 107 = 11235.

Taking R.H.S. = $(a - b)^2 + 4ab$

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$$\begin{aligned}\Rightarrow a^2 - 2ab + b^2 + 4ab\\[1em] \Rightarrow a^2 + 2ab + b^2\\[1em] \Rightarrow (a + b)^2\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\bm{(a + b)^2 = (a - b)^2 + 4ab}$.

Taking L.H.S. = $(2a + 3b)^2 + (2a - 3b)^2$

Using the identities $(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$,

$$\begin{aligned}\Rightarrow [(2a)^2 + 2 \times 2a \times 3b + (3b)^2] + [(2a)^2 - 2 \times 2a \times 3b + (3b)^2]\\[1em] \Rightarrow 4a^2 + 12ab + 9b^2 + 4a^2 - 12ab + 9b^2\\[1em] \Rightarrow (4a^2 + 4a^2) + (12ab - 12ab) + (9b^2 + 9b^2)\\[1em] \Rightarrow 8a^2 + 18b^2\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\bm{(2a + 3b)^2 + (2a - 3b)^2 = 8a^2 + 18b^2}$.

(i) Given,

$x + \frac{1}{x} = 5$

Squaring both sides,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 = 5^2\\[1em] \Rightarrow x^2 + 2 \times x \times \frac{1}{x} + \frac{1}{x^2} = 25\\[1em] \Rightarrow x^2 + 2 + \frac{1}{x^2} = 25\\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 25 - 2\\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 23\end{aligned}$$

Hence, $\bm{x^2 + \frac{1}{x^2} = 23}$.

(ii) From part (i), $x^2 + \frac{1}{x^2} = 23$

Squaring both sides,

$$\begin{aligned}\Rightarrow \Big(x^2 + \frac{1}{x^2}\Big)^2 = 23^2\\[1em] \Rightarrow (x^2)^2 + 2 \times x^2 \times \frac{1}{x^2} + \Big(\frac{1}{x^2}\Big)^2 = 529\\[1em] \Rightarrow x^4 + 2 + \frac{1}{x^4} = 529\\[1em] \Rightarrow x^4 + \frac{1}{x^4} = 529 - 2\\[1em] \Rightarrow x^4 + \frac{1}{x^4} = 527\end{aligned}$$

Hence, $\bm{x^4 + \frac{1}{x^4} = 527}$.

Given,

$a + b = 5$ and $a^2 + b^2 = 13$

We know that $(a + b)^2 = a^2 + b^2 + 2ab$

Substituting the values,

$$\begin{aligned}\Rightarrow (5)^2 = 13 + 2ab\\[1em] \Rightarrow 25 = 13 + 2ab\\[1em] \Rightarrow 2ab = 25 - 13\\[1em] \Rightarrow 2ab = 12\\[1em] \Rightarrow ab = 6\end{aligned}$$

Hence, $\bm{ab = 6}$.

Consecutive odd integers differ by 2.

Given, the smallest odd integer = $2x - 1$

So, the next odd integer = $2x - 1 + 2 = 2x + 1$

And the third odd integer = $2x + 1 + 2 = 2x + 3$

Required polynomial = $(2x - 1)(2x + 1)(2x + 3)$

= $[(2x - 1)(2x + 1)](2x + 3)$

Using the identity $(a - b)(a + b) = a^2 - b^2$,

$$\begin{aligned}\Rightarrow [(2x)^2 - (1)^2](2x + 3)\\[1em] \Rightarrow (4x^2 - 1)(2x + 3)\\[1em] \Rightarrow 4x^2(2x + 3) - 1(2x + 3)\\[1em] \Rightarrow 8x^3 + 12x^2 - 2x - 3\end{aligned}$$

Hence, the required polynomial = $\bm{8x^3 + 12x^2 - 2x - 3}$.

$(a + b + c)^2 = [(a + b) + c]^2$

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$, taking $(a + b)$ as the first term and $c$ as the second term,

$$\begin{aligned}\Rightarrow (a + b)^2 + 2(a + b)c + c^2\\[1em] \Rightarrow a^2 + 2ab + b^2 + 2ac + 2bc + c^2\\[1em] \Rightarrow a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\end{aligned}$$

Hence, $\bm{(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca}$.

Now, for $(2x - 3y + 4z)^2$, we take $a = 2x, b = -3y$ and $c = 4z$.

$$\begin{aligned}\Rightarrow (2x - 3y + 4z)^2\\[1em] \Rightarrow (2x)^2 + (-3y)^2 + (4z)^2 + 2(2x)(-3y) + 2(-3y)(4z) + 2(4z)(2x)\\[1em] \Rightarrow 4x^2 + 9y^2 + 16z^2 - 12xy - 24yz + 16zx\end{aligned}$$

Hence, $\bm{(2x - 3y + 4z)^2 = 4x^2 + 9y^2 + 16z^2 - 12xy - 24yz + 16zx}$.

Formula for $\bm{(a + b)^3}$:

$(a + b)^3 = (a + b)(a + b)^2$

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$,

$$\begin{aligned}\Rightarrow (a + b)(a^2 + 2ab + b^2)\\[1em] \Rightarrow a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2)\\[1em] \Rightarrow a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3\\[1em] \Rightarrow a^3 + (2a^2b + a^2b) + (ab^2 + 2ab^2) + b^3\\[1em] \Rightarrow a^3 + 3a^2b + 3ab^2 + b^3\\[1em] \Rightarrow a^3 + b^3 + 3ab(a + b)\end{aligned}$$

Formula for $\bm{(a - b)^3}$:

$(a - b)^3 = (a - b)(a - b)^2$

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$$\begin{aligned}\Rightarrow (a - b)(a^2 - 2ab + b^2)\\[1em] \Rightarrow a(a^2 - 2ab + b^2) - b(a^2 - 2ab + b^2)\\[1em] \Rightarrow a^3 - 2a^2b + ab^2 - a^2b + 2ab^2 - b^3\\[1em] \Rightarrow a^3 + (-2a^2b - a^2b) + (ab^2 + 2ab^2) - b^3\\[1em] \Rightarrow a^3 - 3a^2b + 3ab^2 - b^3\\[1em] \Rightarrow a^3 - b^3 - 3ab(a - b)\end{aligned}$$

Hence, $\bm{(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3ab(a + b)}$ and $\bm{(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 = a^3 - b^3 - 3ab(a - b)}$.