Algebraic Expressions

The sum of a - b + ab, b - c + bc and c - a + ca is,

= a - b + ab + b - c + bc + c - a + ca

= (a - a) + (- b + b) + ab + bc + ca.

= ab + bc + ca

Hence, option 3 is the correct option.

(x³ - 5x² + 7) + (3x² + 5x - 2) + (2x³ - x + 7)

= (x³ + 2x³) + (-5x² + 3x²) + (5x - x) + (7 - 2 + 7)

= 3x³ - 2x² + 4x + 12

Hence, option 1 is the correct option.

(x³ - 5x² + 3x + 2) - (6x² - 4x³ + 3x + 5)

= x³ - 5x² + 3x + 2 - 6x² + 4x³ - 3x - 5

= (x³ + 4x³) + (- 5x² - 6x²) (+ 3x - 3x) + (2 - 5)

= 5x³ - 11x² - 3

Hence, option 3 is the correct option.

p - (p - q) - q - (q - p)

= p - p + q - q - q + p

= (p - p + p) + (q - q - q)

= p - q

Hence, option 2 is the correct option.

(ab - bc) - (ca - bc) + (ca - ab)

= ab - bc - ca + bc + ca - ab

= (ab - ab) + (bc - bc) + (- ca + ca)

= 0

Hence, option 4 is the correct option.

Constants = -7, $\sqrt{5}$ and 8 - 5

Variables = 7 + x, 7x + yz, $\sqrt{xy}$, $\frac{3yz}{8}$, 4.5y - 3x, 8 - 5x, 8x - 5y x p and 3y²z ÷ 4x

(i) Terms are 5x² and 3 x ax

Hence, number of terms = 2.

(ii) Terms are ax ÷ 4 and 7

Hence, number of terms = 2.

(iii) Terms are ax , by and y x z

Hence, number of terms = 3.

(iv) Terms are 23 and a x b ÷ 2

Hence, number of terms = 2.

Monomials are those algebraic expression which consist of one term.

xy², 3x x 5y and 3x ÷ 5y are monomials.

Binomials are those algebraic expression which consist of two terms.

8 - 3x is binomial.

Trinomials are those algebraic expression which consist of three terms.

3y² - 5y + 8 and 2y ÷ 7 + 3x - 7 are trinomials

Multinominal are those algebraic expression which consist two or more terms.

8 - 3x, 3y² - 5y + 8, 9x - 3x² + 15x³ - 7, 2y ÷ 7 + 3x - 7 and 4 - ax² + bx + y are multinomials.

(i) Degree = 2
Reason — As the polynomial contains 3 variables, we will find the sum of powers of each term.
The sum of powers of the term xy = 1 + 1 = 2
The sum of powers of the term 7z = 1
∵ Highest sum of powers = 2
∴ Degree of polynomial = 2

(ii) Degree = 3
Reason — As the polynomial contains 1 variable, degree of a polynomial is the highest power of the variable in a polynomial expression.
Highest power of polynomial = 3
∴ Degree of polynomial = 3

(iii) Degree = 8
Reason — As the polynomial contains 1 variable, degree of a polynomial is the highest power of the variable in a polynomial expression.
Highest power of polynomial = 8
∴ Degree of polynomial = 8

(iv) Degree = 3
Reason — As the polynomial contains 3 variables, we will find the sum of powers of each term.
The sum of powers of the term xyz = 1 + 1 + 1 = 3
The powers of the term 3 = 0
∵ Highest sum of powers = 3
∴ Degree of polynomial = 3

(v) Degree = 4
Reason — As the polynomial contains 3 variables, we will find the sum of powers of each term.
The sum of powers of the term xy = 1 + 1 = 2
The sum of powers of the term yz² = 1 + 2 = 3
The sum of powers of the term zx³ = 1 + 3 = 4
∵ Highest sum of powers = 4
∴ Degree of polynomial = 4

(vi) Degree = 12
Reason — As the polynomial contains 3 variables, we will find the sum of powers of each term.
The sum of powers of the term x⁵y⁷ = 5 + 7 = 12
The sum of powers of the term - 8x³y⁸ = 3 + 8 = 11
The sum of powers of the term 10x⁴y⁴z⁴ = 4 + 4 + 4 = 12
∵ Highest sum of powers = 12
∴ Degree of polynomial = 12

Coefficient of ab = 7x.

Coefficient of 7a = bx.

Coefficient of 5x² = 1.

Coefficient of 8 = -ax.

Coefficient of 4xy = -1.

-7x² + 18x² + 3x² - 5x²

Arrange the polynomial with like terms together and combine them.

= (-7x² + 18x²) + 3x² - 5x²

= (11x² + 3x²) - 5x²

= (14x² - 5x²)

= 9x²

Hence, -7x² + 18x² + 3x² - 5x²= 9x²

b²y - 9b²y + 2b²y - 5b²y

Arrange the polynomial with like terms together and combine them.

= (b²y - 9b²y) + 2b²y - 5b²y

= (-8b²y + 2b²y) - 5b²y

= (-6b²y - 5b²y)

= -11b²y

Hence, b²y - 9b²y + 2b²y - 5b²y = -11b²y

abx - 15abx - 10abx + 32abx

Arrange the polynomial with like terms together and combine them.

= (abx - 15abx) - 10abx + 32abx

= (-14abx - 10abx) + 32abx

= (-24abx + 32abx)

= 8abx

Hence, abx - 15abx - 10abx + 32abx = 8abx

7x - 9y + 3 - 3x - 5y + 8

Arrange the polynomial with like terms together and combine them.

(7x - 3x) - 9y + 3 - 5y + 8

= 4x + (- 9y - 5y) + 3 + 8

= 4x - 14y + (3 + 8)

= 4x - 14y + 11

Hence, 7x - 9y + 3 - 3x - 5y + 8 = 4x - 14y + 11

3x² + 5xy - 4y² + x² - 8xy - 5y²

Arrange the polynomial with like terms together and combine them.

= (3x² + x²) + 5xy - 4y² - 8xy - 5y²

= 4x² + (5xy - 8xy) - 4y² - 5y²

= 4x² - 3xy + (- 4y² - 5y²)

= 4x² - 3xy - 9y²

Hence, 3x² + 5xy - 4y² + x² - 8xy - 5y² = 4x² - 3xy - 9y²

5a + 3b + a - 2b + 3a + 5b

Arrange the polynomial with like terms together and combine them.

= (5a + a + 3a) + (3b - 2b + 5b)

= 9a + 6b

Hence, addition of 5a + 3b, a - 2b, 3a + 5b is 9a + 6b.

8x - 3y + 7z + (- 4x + 5y - 4z )+ (- x - y - 2z)

= 8x - 3y + 7z - 4x + 5y - 4z - x - y - 2z

Arrange the polynomial with like terms together and combine them.

= (8x - 4x - x) + (- 3y + 5y - y) + (7z - 4z - 2z)

= 3x + y + z

Hence, the addition of 8x - 3y + 7z, - 4x + 5y - 4z, - x - y - 2z is 3x + y + z.

3b - 7c + 10 + 5c - 2b - 15 + 15 + 12c + b

Arrange the polynomial with like terms together and combine them.

= (3b - 2b + b) + (-7c + 5c + 12c) + (10 - 15 + 15)

= 2b + 10c + 10

Hence, addition of 3b - 7c + 10, 5c - 2b - 15, 15 + 12c + b is 2b + 10c + 10

a - 3b + 3 + 2a + 5 - 3c + 6c - 15 + 6b

Arrange the polynomial with like terms together and combine them.

= (a + 2a) + (- 3b + 6b) + (3 + 5 - 15) + (- 3c + 6c)

= 3a + 3b - 7 + 3c

Hence, the addition of a - 3b + 3, 2a + 5 - 3c, 6c - 15 + 6b is 3a + 3b + 3c - 7

13ab - 9cd - xy + 5xy + 15cd - 7ab + 6xy - 3cd

Arrange the polynomial with like terms together and combine them.

= (13ab - 7ab) + (-9cd + 15cd - 3cd) + (-xy + 5xy + 6xy)

= 6ab + 3cd + 10xy

Hence, the addition of 13ab - 9cd - xy, 5xy, 15cd - 7ab, 6xy - 3cd is 6ab + 3cd + 10xy.

x³ - x²y + 5xy² + y³ + (-x³ - 9xy² + y³) + 3x²y + 9xy²

Arrange the polynomial with like terms together and combine them.

= (x³ - x³) + (- x²y + 3x²y) + (5xy² - 9xy² + 9xy²) + (y³ + y³)

= 0 + 2x²y + 5xy² + 2y³

Hence, the addition of x³ - x²y + 5xy² + y³, -x³ - 9xy² + y³, 3x²y + 9xy² is 2x²y + 5xy² + 2y³.

The total savings = ₹ (4x - 6y) + ₹ (6x + 2y) + ₹ (4y - x) + ₹ (y - 2x)

= 4x - 6y + 6x + 2y + 4y - x + y - 2x

Arrange the polynomial with like terms together and combine them.

= (4x + 6x - x - 2x) + (- 6y + 2y + 4y + y)

= 7x + y

Hence, the total saving is ₹ (7x + y).

3xy² - 4xy²

= -xy²

Hence, 3xy² - 4xy² = - 1xy²

8x²y - (-2x²y + 3xy²)

= 8x²y + 2x²y - 3xy²

Arrange the polynomial with like terms together and combine them.

= (8x²y + 2x²y) - 3xy²

= 10x²y - 3xy²

Hence, 8x²y - (- 2x²y + 3xy²) = 10x²y - 3xy²

7a - 3b + c - 2d - (3a - 5b + c + 2d)

= 7a - 3b + c - 2d - 3a + 5b - c - 2d

Arrange the polynomial with like terms together and combine them.

= (7a - 3a) + (- 3b + 5b) + (c - c) + (- 2d - 2d)

= 4a + 2b - 4d

Hence, 7a - 3b + c - 2d - (3a - 5b + c + 2d) = 4a + 2b - 4d.

3x³ - x² + 6 - (x³ - 4x - 1)

= 3x³ - x² + 6 - x³ + 4x + 1

Arrange the polynomial with like terms together and combine them.

= (3x³ - x³) - x² + 4x + (6 + 1)

= 2x³ - x² + 4x + 7

Hence, 3x³ - x² + 6 - (x³ - 4x - 1) = 2x³ - x² + 4x + 7

a³ - 3a² + 4a + 1 - (6a + 3)

= a³ - 3a² + 4a + 1 - 6a - 3

Arrange the polynomial with like terms together and combine them.

= a³ - 3a² + (4a - 6a) + (1 - 3)

= a³ - 3a² - 2a - 2

Hence, a³ - 3a² + 4a + 1 - (6a + 3) = a³ - 3a² - 2a - 2.

3abc + 5bcd - cda - (cab - 4cad - cbd)

= 3abc + 5bcd - cda - cab + 4cad + cbd

Arrange the polynomial with like terms together and combine them.

= (3abc - abc) + (5bcd + bcd) + (- cda + 4cda)

= 2abc + 6bcd + 3cda

Hence, 3abc + 5bcd - cda - (cab - 4cad - cbd) = 2abc + 6bcd + 3cda.

4a² - 3ab + 2b² - (a² + ab + b²)

= 4a² - 3ab + 2b² - a² - ab - b²

Arrange the polynomial with like terms together and combine them.

= (4a² - a²) + (- 3ab - ab) + (2b² - b²)

= 3a² - 4ab + b²

Hence, 4a² - 3ab + 2b² - (a² + ab + b²) = 3a² - 4ab + b².

3x³ - 4x² + 5x - 6 - (- 3x³ + 4x² - 5x + 6)

= 3x³ - 4x² + 5x - 6 + 3x³ - 4x² + 5x - 6

Arrange the polynomial with like terms together and combine them.

= (3x³ + 3x³) + (- 4x² - 4x²) + (5x + 5x) + (- 6 - 6)

= 6x³ - 8x² + 10x - 12

Hence, 3x³ - 4x² + 5x - 6 - (- 3x³ + 4x² - 5x + 6) = 6x³ - 8x² + 10x - 12.

Difference between m² + m + 4 and - m² + 3m + 6

- m² + 3m + 6 - (m² + m + 4)

= - m² + 3m + 6 - m² - m - 4

Arrange the polynomial with like terms together and combine them.

= (- m² - m²) + (3m - m) + (6 - 4)

= - 2m² + 2m + 2

And, the difference between - 2m² + 2m + 2 and m² + m + 1

(m² + m + 1) - (- 2m² + 2m + 2)

= m² + m + 1 + 2m² - 2m - 2

Arrange the polynomial with like terms together and combine them.

= ( m² + 2m²) + (m - 2m) + (1 - 2)

= 3m² - m - 1

Hence, the result is 3m² - m - 1.

The sum of 5y² + y - 3 and y² - 3y + 7

5y² + y - 3 + (y² - 3y + 7)

= 5y² + y - 3 + y² - 3y + 7

Arrange the polynomial with like terms together and combine them.

= (5y² + y²) + (y - 3y) + (- 3 + 7)

= 6y² - 2y + 4

The difference of 6y² - 2y + 4 from 6y² + y - 2

= 6y² + y - 2 - (6y² - 2y + 4)

= 6y² + y - 2 - 6y² + 2y - 4

= (6y² - 6y²) + (y + 2y) + (- 2 - 4)

= 0 + 3y - 6

Hence, the result is 3y - 6.

Let the number Z must be added to x⁴ - x³ + x² + x + 3 to obtain x⁴ + x² - 1

⇒ Z + (x⁴ - x³ + x² + x + 3) = (x⁴ + x² - 1)

⇒ Z = (x⁴ + x² - 1) - (x⁴ - x³ + x² + x + 3)

⇒ Z = x⁴ + x² - 1 - x⁴ + x³ - x² - x - 3

Arrange the polynomial with like terms together and combine them.

⇒ Z = (x⁴ - x⁴) + x³ + (x² - x²) - x + (- 1 - 3)

⇒ Z = 0 + x³ + 0 - x - 4

Hence, the number is x³ - x - 4.

Let 5x² + 10xy - y² be more than 2x² + 4xy + 2y² by Z.

2x² + 4xy + 2y² + Z = 5x² + 10xy - y²

⇒ Z = 5x² + 10xy - y² - (2x² + 4xy + 2y²)

⇒ Z = 5x² + 10xy - y² - 2x² - 4xy - 2y²

Arrange the polynomial with like terms together and combine them.

⇒ Z = (5x² - 2x²) + (10xy - 4xy) + (- y² - 2y²)

⇒ Z = 3x² + 6xy - 3y²

Hence, the number is 3x² + 6xy - 3y².

Let 2a² + 1 be less than 3a² - 6 by Z.

3a² - 6 - Z = 2a² + 1

⇒ Z = 3a² - 6 - (2a² + 1)

⇒ Z = 3a² - 6 - 2a² - 1

Arrange the polynomial with like terms together and combine them.

⇒ Z = (3a² - 2a²) + (- 6 - 1)

⇒ Z = a² - 7

Hence, the number is a² - 7.

(i) If x = 6a + 8b + 9c; y = 2b - 3a - 6c and z = c - b + 3a, then

x + y + z = (6a + 8b + 9c) + (2b - 3a - 6c) + (c - b + 3a)

= 6a + 8b + 9c + 2b - 3a - 6c + c - b + 3a

Arrange the polynomial with like terms together and combine them.

= (6a - 3a + 3a) + (8b + 2b - b) + (9c - 6c + c)

= 6a + 9b + 4c

Hence, x + y + z = 6a + 9b + 4c

(ii) If x = 6a + 8b + 9c; y = 2b - 3a - 6c and z = c - b + 3a, then

x - y + z = (6a + 8b + 9c) - (2b - 3a - 6c) + (c - b + 3a)

= 6a + 8b + 9c - 2b + 3a + 6c + c - b + 3a

Arrange the polynomial with like terms together and combine them.

= (6a + 3a + 3a) + (8b - 2b - b) + (9c + 6c + c)

= 12a + 5b + 16c

Hence, x - y + z = 12a + 5b + 16c

(iii) If x = 6a + 8b + 9c; y = 2b - 3a - 6c and z = c - b + 3a, then

2x - y - 3z = 2(6a + 8b + 9c) - (2b - 3a - 6c) - 3(c - b + 3a)

= 12a + 16b + 18c - 2b + 3a + 6c - 3c + 3b - 9a

Arrange the polynomial with like terms together and combine them.

= (12a + 3a - 9a) + (16b - 2b + 3b) + (18c + 6c - 3c)

= 6a + 17b + 21c

Hence, 2x - y - 3z = 6a + 17b + 21c

(iv) If x = 6a + 8b + 9c; y = 2b - 3a - 6c and z = c - b + 3a, then

3y - 2z - 5x = 3(2b - 3a - 6c) - 2(c - b + 3a) - 5(6a + 8b + 9c)

= 6b - 9a - 18c - 2c + 2b - 6a - 30a - 40b - 45c

Arrange the polynomial with like terms together and combine them.

= (-9a - 6a - 30a) + (6b + 2b - 40b) + (-18c - 2c - 45c)

= -45a - 32b - 65c

Hence, 3y - 2z - 5x = -45a - 32b - 65c

(3x) $\times$ (9x⁴ - 8x³ - 12x)

= 3x $\times$ 9x⁴ - 3x $\times$ 8x³ - 3x $\times$ 12x

= 27x⁽⁴⁺¹⁾ - 24x⁽³⁺¹⁾ - 36x⁽¹⁺¹⁾

= 27x^5 - 24x^4 - 36x^2

Hence, option 2 is the correct option.

(9x⁴ - 12x³ - 18x) ÷ (3x)

= (9x⁴ ÷ 3x) - (12x³ ÷ 3x) - (18x ÷ 3x)

= 3x^(4-1) - 4x^(3-1) - 6x^(1-1)

= 3x³ - 4x² - 6

Hence, option 3 is the correct option.

$$\begin{aligned}\Big(\frac{10}{3}xy^2z\Big) \times \Big(-\frac{9}{5}x^2z\Big)\\[1em] = \Big(-\frac{10 \times 9}{3 \times 5}x^{1+2}y^2z^{1+1}\Big)\\[1em] = \Big(-\frac{90}{15}x^{3}y^2z^{2}\Big)\\[1em] = (-6x^3y^2z^2)\end{aligned}$$

Hence, option 1 is the correct option.

(x³+ y²) $\times$ 10x²

= 10x² $\times$ x³ + 10x² $\times$ y²

= 10x⁽²⁺³⁾ + 10x²y²

= 10x⁵ + 10x²y²

Hence, option 2 is the correct option.

(a³ - b³) ÷ (a - b)

[∵ As we know, (x³ - y³) = (x - y)(x² + y² + xy)]

∴ (a³ - b³) ÷ (a - b) = $\frac{(a - b)(a^2 + b^2 + ab)}{(a - b)}$

= $\frac{\cancel{(a - b)}(a^2 + b^2 + ab)}{\cancel{(a - b)}}$

= (a² + b² + ab)

Hence, option 1 is the correct option.

8ab² x (- 4a³b⁴)

= 8 x (-4) a⁽¹⁺³⁾ b⁽²⁺⁴⁾

= -32 a⁴ b⁶

Hence, 8ab² x (- 4a³b⁴) = -32a⁴ b⁶

$$\begin{aligned}\frac{2}{3}ab \times \Big(-\frac{1}{4}a^2b\Big)\\[1em] = -\frac{2 \times 1}{3 \times 4}a^{1+2}b^{1+1}\\[1em] = -\frac{2}{12}a^3b^2\\[1em] = -\frac{1}{6}a^3b^2\end{aligned}$$

Hence, $\frac{2}{3}$ ab x (- $\frac{1}{4}$ a²b) = $-\frac{1}{6}$a³b².

(-5cd²) x (- 5cd²)

= (-5) x (-5) c⁽¹⁺¹⁾d⁽²⁺²⁾

= 25c²d⁴

Hence, (-5cd²) x (-5cd²) = 25c²d⁴

4a x (6a + 7)

= 4a x 6a + 4a x 7

= 24a⁽¹⁺¹⁾ + 28a

= 24a² + 28a

Hence, 4a x (6a + 7) = 24a² + 28a

-8x $\times$ (4 - 2x - x²)

= (-8x) $\times$ 4 - (-8x) $\times$ 2x - (-8x) $\times$ x²

= -32x + 16x⁽¹⁺¹⁾ + 8x⁽¹⁺²⁾

= -32x + 16x² + 8x³

Hence, (-8x) x (4 - 2x - x²) = -32x + 16x² + 8x³

-3a x (2a² - 5a - 4)

= -3a x 2a² - (-3a) x 5a - (-3a) x 4

= -6a⁽²⁺¹⁾ + 15a⁽¹⁺¹⁾ + 12a

= -6a³ + 15a² + 12a

Hence, (2a² - 5a - 4) x (-3a) = -6a³ + 15a² + 12a

(x + 4) $\times$ (x - 5)

= x $\times$ (x - 5) + 4 $\times$ (x - 5)

= x $\times$ x - x $\times$ 5 + 4 $\times$ x - 4 $\times$ 5

= x² - 5x + 4x - 20

= x² - x - 20

Hence, (x + 4) x (x - 5) = x² - x - 20

(5a - 1) x (7a - 3)

= 5a x (7a - 3) - 1 x (7a - 3)

= 5a x 7a - 5a x 3 - 1 x 7a - 1 x (-3)

= 35a⁽¹⁺¹⁾ - 15a - 7a + 3

= 35a² - 22a + 3

Hence, (5a - 1) x (7a - 3) = 35a² - 22a + 3

(12a + 5b) x (7a - b)

= 12a x 7a - 12a x b + 5b x 7a - 5b x b

= 84a⁽¹⁺¹⁾ - 12ab + 35ab - 5b⁽¹⁺¹⁾

= 84a² + 23ab - 5b²

Hence, (12a + 5b) x (7a - b) = 84a² + 23ab - 5b²

(x² + x + 1) $\times$ (1 - x)

= 1 $\times$ (x² + x + 1) - x $\times$ (x² + x + 1)

= (1 $\times$ x² + 1 $\times$ x + 1 $\times$ 1) - (x $\times$ x² + x $\times$ x + x $\times$ 1)

= x² + x + 1 - (x³ + x² + x)

= x² + x + 1 - x³ - x² - x

= (x² - x²) + (x - x) + 1 - x³

= 1 - x³

Hence, (x² + x + 1) $\times$ (1 - x) = 1 - x³

(2m² - 3m - 1) x (4m² - m - 1)

= 2m² x 4m² - 2m² x m - 2m² x 1 - 3m x 4m² - 3m x (-m) - 3m x (-1) - 1 x 4m² - 1 x (- m) - 1 x (-1)

= 8m⁽²⁺²⁾ - 2m⁽²⁺¹⁾ - 2m² - 12m⁽²⁺¹⁾ + 3m⁽¹⁺¹⁾ + 3m - 4m² + m + 1

= 8m⁴ - 2m³ - 2m² - 12m³ + 3m² + 3m - 4m² + m + 1

= 8m⁴ - 2m³ - 12m³ - 2m² + 3m² - 4m² + 3m + m + 1

= 8m⁴ - 14m³ - 3m² + 4m + 1

Hence, (2m² - 3m - 1) x (4m² - m - 1) = 8m⁴ - 14m³ - 3m² + 4m + 1

(a² x ab) x b²

= a⁽²⁺¹⁾b x b²

= a³b x b²

= a³b⁽²⁺¹⁾

= a³b³

Hence, a² x ab x b² = a³b³

abx $\times$ (- 3a²x) $\times$ 7b²x³

= (-3 $\times$ 7) a⁽¹⁺²⁾b⁽¹⁺²⁾x^(1+1+3)

= -21 a³b³x⁵

Hence, (abx) $\times$ (- 3a²x) $\times$ (7b²x³) = -21a³b³x⁵

-3bx $\times$ (-5xy) $\times$ (-7b³y²)

= -3 $\times$ (-5) $\times$ (-7) b⁽¹⁺³⁾x⁽¹⁺¹⁾y⁽¹⁺²⁾

= -105 b⁴x²y³

Hence, (- 3bx) $\times$ (- 5xy) $\times$ (- 7b³y²) = -105b⁴x²y³

$$\begin{aligned}-\frac{3}{2}x^5y^3 \times \frac{4}{9}a^2x^3y\\[1em] = -\frac{3 \times 4}{2 \times 9}a^2x^{5+3}y^{3+1}\\[1em] = -\frac{12}{18}a^2x^8y^4\\[1em] = -\frac{2}{3}a^2x^8y^4\\[1em]\end{aligned}$$

Hence, - $\frac{3}{2}$ x⁵ y³ x $\frac{4}{9}$ a²x³y = - $\frac{2}{3}$ a²x⁸y⁴

$$\begin{aligned}-\frac{2}{3}a^7b^2 \times \Big(-\frac{9}{4}ab^5\Big)\\[1em] = \frac{2 \times 9}{3 \times 4}a^{7 + 1}b^{2+5}\\[1em] = \frac{18}{12}a^8b^7\\[1em] = \frac{3}{2}a^8b^7\\[1em]\end{aligned}$$

Hence, - $\frac{2}{3}$ a⁷b² x - $\frac{9}{4}$ ab⁵ = $\frac{3}{2}$ a⁸b⁷

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

Hence, 2a³ - 3a²b x $-\frac{1}{2}$ ab² = -a⁴b² + $\frac{3}{2}$ a³b³

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

Hence, 2x + $\frac{1}{2}$ y x 2x - $\frac{1}{2}$ y = 4x² - $\frac{1}{4}$ y²

(5x² - 8xy + 6y² - 3) $\times$ (- 3xy)

= 5x² $\times$ (- 3xy) - 8xy $\times$ (- 3xy) + 6y² $\times$ (- 3xy) - 3 $\times$ (- 3xy)

= -15x⁽²⁺¹⁾y + 24x⁽¹⁺¹⁾y⁽¹⁺¹⁾ - 18xy⁽²⁺¹⁾ + 9xy

= -15x³y + 24x²y² - 18xy³ + 9xy

Hence, (5x² - 8xy + 6y² - 3) x (- 3xy) = -15x³y + 24x²y² - 18xy³ + 9xy

$$\begin{aligned}\Big(3 - \frac{2}{3}xy + \frac{5}{7}xy^2 - \frac{16}{21}x^2y\Big) \times (-21x^2y^2)\\[1em] = \Big(3 \times (-21x^2y^2) - \frac{2}{3}xy \times (-21x^2y^2) + \frac{5}{7}xy^2 \times (-21x^2y^2) - \frac{16}{21}x^2y \times (-21x^2y^2)\Big)\\[1em] = \Big(-63x^2y^2 - \frac{2 \times (-21)}{3}x^{1+2}y^{1+2} + \frac{5\times (-21)}{7}x^{1+2}y^{2+2} - \frac{16 \times (-21)}{21}x^{2+2}y^{1+2}\Big)\\[1em] = -63x^2y^2 - \frac{-42}{3}x^3y^3 + \frac{-105}{7}x^3y^4 + \frac{16 \times \cancel{(21)}}{\cancel{21}}x^4y^3\\[1em] = -63x^2y^2 + 14x^3y^3 - 15x^3y^4 + 16x^4y^3\end{aligned}$$

Hence, (3 - $\frac{2}{3} xy + \frac{5}{7} xy^2 - \frac{16}{21}$ x²y) x (- 21x²y²) = -63x²y² + 14x³y³ - 15x³y⁴ + 16x⁴y³

(6x³ - 5x + 10) $\times$ (4 - 3x²)

= 4 $\times$ (6x³ - 5x + 10) - 3x² $\times$ (6x³ - 5x + 10)

= 24x³ - 20x + 40 - 18x⁽³⁺²⁾ + 15x⁽¹⁺²⁾ - 30x²

= 24x³ - 20x + 40 - 18x⁵ + 15x³ - 30x²

= - 18x⁵ + 24x³ + 15x³ - 30x² - 20x + 40

= - 18x⁵ + 39x³ - 30x² - 20x + 40

Hence, (6x³ - 5x + 10) x (4 - 3x²) = - 18x⁵ + 39x³ - 30x² - 20x + 40

(2y - 4y³ + 6y⁵) x (y² + y - 3)

= 2y x (y² + y - 3) - 4y³ x (y² + y - 3) + 6y⁵ x (y² + y - 3)

= 2y⁽¹⁺²⁾ + 2y⁽¹⁺¹⁾ - 6y - 4y⁽³⁺²⁾ - 4y⁽³⁺¹⁾ + 12y³ + 6y⁽⁵⁺²⁾ + 6y⁽⁵⁺¹⁾ - 18y⁵

= 2y³ + 2y² - 6y - 4y⁵ - 4y⁴ + 12y³ + 6y⁷ + 6y⁶ - 18y⁵

= 6y⁷ + 6y⁶ - 18y⁵ - 4y⁵ - 4y⁴ + 12y³ + 2y³ + 2y² - 6y

= 6y⁷ + 6y⁶ - 22y⁵ - 4y⁴ + 14y³ + 2y² - 6y

Hence, (2y - 4y³ + 6y⁵) x (y² + y - 3) = 6y⁷ + 6y⁶ - 22y⁵ - 4y⁴ + 14y³ + 2y² - 6y

(5p² + 25pq + 4q²) x (2p² - 2pq + 3q²)

= 5p² x (2p² - 2pq + 3q²) + 25pq x (2p² - 2pq + 3q²) + 4q² x (2p² - 2pq + 3q²)

= 10p⁽²⁺²⁾ - 10p⁽²⁺¹⁾q + 15p²q² + 50p⁽¹⁺²⁾q - 50p⁽¹⁺¹⁾q⁽¹⁺¹⁾ + 75pq⁽¹⁺²⁾ + 8p²q² - 8pq⁽²⁺¹⁾ + 12q⁽²⁺²⁾

= 10p⁴ - 10p³q + 15p²q² + 50p³q - 50p²q² + 75pq³ + 8p²q² - 8pq³ + 12q⁴

= 10p⁴ - 10p³q + 50p³q + 15p²q² - 50p²q² + 8p²q² + 75pq³ - 8pq³ + 12q⁴

= 10p⁴ + 40p³q - 27p²q² + 67pq³ + 12q⁴

Hence, (5p² + 25pq + 4q²) x (2p² - 2pq + 3q²) = 10p⁴ + 40p³q - 27p²q² + 67pq³ + 12q⁴

(7x - 8) (3x + 2)

= 7x (3x + 2) - 8 (3x + 2)

= 21x⁽¹⁺¹⁾ + 14x - 24x - 16

= 21x² + 14x - 24x - 16

= 21x² - 10x - 16

Hence, (7x - 8) (3x + 2) = 21x² - 10x - 16

(px - q) (px + q)

= px (px + q) - q (px + q)

= p⁽¹⁺¹⁾x⁽¹⁺¹⁾ + pqx - pqx - q⁽¹⁺¹⁾

= p²x² + pqx - pqx - q²

= p²x² - q²

Hence, (px - q) (px + q) = p²x² - q²

(5a + 5b - c) (2b - 3c)

= 5a (2b - 3c) + 5b (2b - 3c) - c (2b - 3c)

= 10ab - 15ac + 10b⁽¹⁺¹⁾ - 15bc - 2bc + 3c⁽¹⁺¹⁾

= 10ab - 15ac + 10b² - 17bc + 3c²

Hence, (5a + 5b - c) (2b - 3c) = 10ab - 15ac + 10b² - 17bc + 3c²

(4x - 5y) (5x - 4y)

= 4x (5x - 4y) - 5y (5x - 4y)

= 20x⁽¹⁺¹⁾ - 16xy - 25xy + 20y⁽¹⁺¹⁾

= 20x² - 41xy + 20y²

Hence, (4x - 5y) (5x - 4y) = 20x² - 41xy + 20y²

(3y + 4z) (3y - 4z) + (2y + 7z) (y + z)

= [3y(3y - 4z) + 4z(3y - 4z)] + [2y (y + z) + 7z (y + z)]

= [9y² - 12yz + 12yz - 16z²] + [2y² + 2yz + 7zy + 7z²]

= 9y² - 16z² + 2y² + 9zy + 7z²

= 9y² + 2y² + 9zy - 16z² + 7z²

= 11y² + 9yz - 9z²

Hence, (3y + 4z) (3y - 4z) + (2y + 7z) (y + z) = 11y² + 9yz - 9z²

Given:

Dimensions of the rectangle = x² - 4xy + 7y² and x³ - 5xy²

As we know, area of rectangle = l x b

So, putting the value

(x² - 4xy + 7y²) $\times$ (x³ - 5xy²)

= x² $\times$ (x³ - 5xy²) - 4xy $\times$ (x³ - 5xy²) + 7y² $\times$ (x³ - 5xy²)

= x⁽²⁺³⁾ - 5x⁽²⁺¹⁾y² - 4x⁽¹⁺³⁾y + 20x⁽¹⁺¹⁾y⁽¹⁺²⁾ + 7x³y² - 35xy⁽²⁺²⁾

= x⁵ - 5x³y² + 7x³y² - 4x⁴y + 20x²y³ - 35xy⁴

= x⁵ + 2x³y² - 4x⁴y + 20x²y³ - 35xy⁴

Hence, the area of the rectangle = x⁵ + 2x³y² - 4x⁴y + 20x²y³ - 35xy⁴

Given:

Base of a triangle = (3x - 4y)

Altitude of a triangle = (6x + 5y)

As we know that the area of triangle = $\frac{1}{2}$ x base x altitude

$$\begin{aligned}\frac{1}{2} \times (3x - 4y) \times (6x + 5y)\\[1em] = \frac{1}{2} \times (3x (6x + 5y) - 4y (6x + 5y))\\[1em] = \frac{1}{2} \times (18x^2 + 15xy - 24xy - 20y^2)\\[1em] = \frac{1}{2} \times (18x^2 - 9xy - 20y^2)\end{aligned}$$

Hence, the area of tirangle = $\frac{1}{2}$ (18x² - 9xy - 20y²) sq. unit

(- 4xy³) $\times$ (6x²y)

= - 24x⁽¹⁺²⁾y⁽³⁺¹⁾

= - 24x³y⁴

Hence, (- 4xy³) $\times$ (6x²y) = - 24x³y⁴

Putting x = 2 and y = 1

Taking LHS:

(- 4xy³) x (6x²y)

= (- 4 x 2 x 1³) x (6 x 2² x 1)

= (- 4 x 2 x 1) x (6 x 4 x 1)

= (- 8) x (24)

= - 192

Taking RHS:

- 24x³y⁴

= - 24 x 2³ x 1⁴

= - 24 x 8 x 1

= - 192

Hence, LHS = RHS

So, (- 4xy³) $\times$ (6x²y) = - 24x³y⁴

(3x³) $\times$ (-5xy²) $\times$ (2x²yz³)

= (- 15x⁽³⁺¹⁾y²) $\times$ (2x²yz³)

= (- 15x⁴y²) $\times$ (2x²yz³)

= - 30x⁽⁴⁺²⁾y⁽²⁺¹⁾z³

= - 30x⁶y³z³

Putting the value x = 1, y = 2 and z = 3, we get

- 30 x 1⁶ x 2³ x 3³

= - 30 x 1 x 8 x 27

= - 6480

Hence, the value of (3x³) x (-5xy²) x (2x²yz³) for x = 1, y = 2 and z = 3 is - 6480

(3x⁴y²) $\times$ (2x²y³)

= 6x⁽⁴⁺²⁾y⁽²⁺³⁾

= 6x⁶y⁵

Putting x = 1 and y = 2, we get

6 x 1⁶ x 2⁵

= 6 x 1 x 32

= 192

Hence, (3x⁴y²) (2x²y³) for x = 1 and y = 2 is 192

Given:

(x⁵) $\times$ (3x²) $\times$ (-2x)

= 3x⁽⁵⁺²⁾ $\times$ (-2x)

= 3x⁷ $\times$ (-2x)

= - 6x⁽⁷⁺¹⁾

= - 6x⁸

For x = 1

- 6 x 1⁸

= - 6 x 1

= - 6

Hence, the value of (x⁵) $\times$ (3x²) x (-2x) for x = 1 is -6

- 70a³ ÷ 14a²

= - $\frac{70}{14}$ a^(3-2)

= - 5 a¹

Hence, - 70a³ ÷ 14a² = - 5 a

(24x³y³) ÷ (- 8y²)

= - $\frac{24}{8}$ x³y^(3-2)

= - 3 x³y¹

Hence, (24x³y³) ÷ (- 8y²) = - 3 x³y

(15a⁴b) ÷ (- 5a³b)

= - $\frac{15}{5}$ a^(4-3)b^(1-1)

= - 3a¹b⁰

Hence, (15a⁴b) ÷ (- 5a³b) = - 3a

(- 24x⁴d³) ÷ (- 2x²d⁵)

= $\frac{24}{2}$ x^(4-2)d^(3-5)

= 12 x²d^(-2)

= $\frac{12x^2}{d^2}$

Hence, (- 24x⁴d³) ÷ (- 2x²d⁵) = $\frac{12x^2}{d^2}$

(63a⁴b⁵c⁶) ÷ (- 9a²b⁴c³)

= - $\frac{63}{9}$ a^(4-2)b^(5-4)c^(6-3)

= - 7a²b¹c³

Hence, (63a⁴b⁵c⁶) ÷ (- 9a²b⁴c³) = - 7a²bc³

(8x - 10y + 6c) ÷ 2

= (8x ÷ 2) - (10y ÷ 2) + (6c ÷ 2)

= 4x - 5y + 3c

Hence, (8x - 10y + 6c) ÷ 2 = 4x - 5y + 3c

(15a³b⁴ - 10a⁴b³ - 25a³b⁶) ÷ (-5a³b²)

= (15a³b⁴ ÷ (- 5a³b²)) - (10a⁴b³ ÷ (- 5a³b²)) - (25a³b⁶ ÷ (- 5a³b²))

= $\Big($- $\frac{15}{5}$ a^(3-3)b^(4-2)$\Big)$ - $\Big($- $\frac{10}{5}$ a^(4-3)b^(3-2)$\Big)$ - $\Big($- $\frac{25}{5}$ a^(3-3)b^(6-2)$\Big)$

= - 3 a⁰b² + 2 a¹b¹ + 5 a⁰b⁴

= - 3b² + 2 ab + 5b⁴

Hence, (15a³b⁴ - 10a⁴b³ - 25a³b⁶) ÷ (- 5a³b²) = - 3b² + 2 ab + 5b⁴

(- 14x⁶y³ - 21x⁴y⁵ + 7x⁵y⁴) ÷ 7x²y²

= (- 14x⁶y³ ÷ 7x²y²) - (21x⁴y⁵ ÷ 7x²y²) + (7x⁵y⁴ ÷ 7x²y²)

= $\Big($ - $\frac{14}{7}$ x^(6-2)y^(3-2) $\Big)$ - $\Big($ $\frac{21}{7}$ x^(4-2)y^(5-2) $\Big)$ - $\Big($ - $\frac{7}{7}$ x^(5-2)y^(4-2) $\Big)$

= - 2 x⁴y¹ - 3 x²y³ + 1 x³y²

Hence, (- 14x⁶y³ - 21x⁴y⁵ + 7x⁵y⁴) ÷ 7x²y² = - 2 x⁴y - 3 x²y³ + x³y²

(a² + 7a + 12) ÷ (a + 4)

= (a² + 3a + 4a + 12) ÷ (a + 4)

= [a(a + 3) + 4(a + 3)] ÷ (a + 4)

= [(a + 3)(a + 4)] ÷ (a + 4)

= $\frac{(a + 3)\cancel{(a + 4)}}{\cancel{(a + 4)}}$

= (a + 3)

Hence, (a² + 7a + 12) ÷ (a + 4) = a + 3

(x² + 3x - 54) ÷ (x - 6)

= (x² + 9x - 6x - 54) ÷ (x - 6)

= [x(x + 9) - 6(x + 9)] ÷ (x - 6)

= [(x + 9)(x - 6)] ÷ (x - 6)

= $\frac{(x + 9)\cancel{(x - 6)}}{\cancel{(x - 6)}}$

= (x + 9)

Hence, (x² + 3x - 54) ÷ (x - 6) = x + 9

(12x² + 7xy - 12y²) ÷ (3x + 4y)

= (12x² + 16xy - 9xy - 12y²) ÷ (3x + 4y)

= [4x(3x + 4y) - 3y(3x + 4y)] ÷ (3x + 4y)

= [(3x + 4y)(4x - 3y)] ÷ (3x + 4y)

= $\frac{[\cancel{(3x + 4y)}(4x - 3y)]}{\cancel{(3x + 4y)}}$

= 4x - 3y

Hence, (12x² + 7xy - 12y²) ÷ (3x + 4y) = (4x - 3y)

(x⁶ - 8) ÷ (x² - 2)

= (x^2x3 - 2³) ÷ (x² - 2)

[Using the formula (a³ - b³) = (a - b)(a² + ab + b²)]

= [(x² - 2)(x⁴ + 2x² + 4)] ÷ (x² - 2)

= $\frac{\cancel{(x^2 - 2)}(x^4 + 2x^2 + 4)}{\cancel{(x^2 - 2)}}$

= (x⁴ + 2x² + 4)

Hence,(x⁶ - 8) ÷ (x² - 2) = (x⁴ + 2x² + 4)

Dividing 6x³ - 13x² - 13x + 30 by 2x² - x - 6

$$\begin{array}{l} \phantom{2x^2 - x - 6)}{3x - 5} \\ 2x^2 - x - 6\overline{\smash{\big)}6x^3 - 13x^2 - 13x + 30} \\ \phantom{2x^2 - x - 6}\underline{\underset{-}{}6x^3 \underset{+}{-}\phantom{0}3x^2 \underset{+}{-}18x} \\ \phantom{{2x^2 - x - 6}2x^3+}-10x^2 + 5x + 30 \\ \phantom{{2x^2 - x - 6}2x^3+}\underline{\underset{+}{-}10x^2 \underset{-}{+} 5x \underset{-}{+} 30} \\ \phantom{{2x^2 - x - 6}{2x^3+}{+5x^2222}}\times \end{array}$$

Hence, (6x³ - 13x² - 13x + 30) ÷ (2x² - x - 6) = 3x - 5

Dividing 4a² + 12ab + 9b² - 25c² by 2a + 3b + 5c

$$\begin{array}{l} \phantom{2a + 3b + 5c)}{2a + 3b - 5c} \\ 2a + 3b + 5c\overline{\smash{\big)}4a^2 + 12ab + 9b^2 - 25c^2} \\ \phantom{2a + 3b + 5c}\underline{\underset{-}{}4a^2 \underset{-}{+}6ab \phantom{+ 9b^2 - 25c^2} \underset{-}{+}10ac} \\ \phantom{{2a + 3b + 5c}2x^3+}6ab - 10ac + 9b^2 - 25c^2 \\ \phantom{{2a + 3b + 5c}2x+}\underline{\underset{-}{}6ab \phantom{- 10ac} \underset{-}{+} 9b^2 \phantom{- 25c^2} \underset{-}{+} 15bc} \\ \phantom{{2a + 3b + 5c}{2x^3+}{+5x^2+}}-10ac - 25c^2 -15bc \\ \phantom{{2a + 3b + 5c}{2x^3+}{+5x^2+\enspace}}\underline{\underset{+}{-}10ac \underset{+}{-} 25c^2 \underset{+}{-} 15bc} \\ \phantom{{2a + 3b + 5c}{2x^3+}{+5x^2+\qquad}{-23x}}\times \end{array}$$

Hence, (4a² + 12ab + 9b² - 25c²) ÷ (2a + 3b + 5c) = 2a + 3b - 5c

Dividing 16 + 8x + x⁶ - 8x³ - 2x⁴ + x² by x + 4 - x³

⇒ Dividing x⁶ - 2x⁴ - 8x³ + x² + 8x + 16 by -x³ + x + 4

$$\begin{array}{l} \phantom{-x^3 + x + 4)}{-x^3 + x + 4} \\ -x^3 + x + 4\overline{\smash{\big)}x^6 - 2x^4 - 8x^3 + x^2 + 8x + 16} \\ \phantom{-x^3 + x + 4}\underline{\underset{-}{}x^6 \underset{+}{-}\phantom{2}x^4 \underset{+}{-}4x^3} \\ \phantom{{-x^3 + x + 4}2x^3+}-x^4 - 4x^3 + x^2 + 8x + 16 \\ \phantom{{-x^3 + x + 4}2x+2}\underline{\underset{+}{-}x^4 \phantom{- 4x^3} \underset{-}{+} x^2 \underset{-}{+} 4x} \\ \phantom{{-x^3 + x + 4}{2x^3+}{+x+}}-4x^3 \phantom{+ x^2} + 4x + 16 \\ \phantom{{-x^3 + x + 4}{2x^3+}{+541)}}\underline{\underset{+}{-}4x^3 \phantom{+ x^2} \underset{+}{-} 4x \underset{-}{+} 16} \\ \phantom{{-x^3 + x + 4}{2x^3+}{+5x^2+\qquad}{-23x}}\times \end{array}$$

Hence, (16 + 8x + x⁶ - 8x³ - 2x⁴ + x²) ÷ (x + 4 - x³) = -x³ + x + 4

Dividing a³ - 5a² + 8a + 15 by a + 1

$$\begin{array}{l} \phantom{a + 1)}{a^2 - 6a + 14} \\ a + 1\overline{\smash{\big)}a^3 - 5a^2 + 8a + 15} \\ \phantom{a + 1}\underline{\underset{-}{}a^3 \underset{-}{+}a^2} \\ \phantom{{a + 1}2x}-6a^2 +8a + 15 \\ \phantom{{a + 1}21x}\underline{\underset{+}{-}6a^2 \underset{+}{-} 6a} \\ \phantom{{a + 1}{2x^3+}{+5x^2+}}14a + 15 \\ \phantom{{a + 1}{2x^3+}{+5x^2}}\underline{\underset{+}{-}14a \underset{-}{+} 14} \\ \phantom{{a + 1}{a^3 - 5a^2 + 8a + 152}} 1 \end{array}$$

Quotient = a² - 6a + 14

Remainder = 1

Verification:

Quotient x Divisor + Remainder

= (a² - 6a + 14) $\times$ (a + 1) + 1

= a $\times$ (a² - 6a + 14) + 1 $\times$ (a² - 6a + 14) + 1

= a⁽¹⁺²⁾ - 6a⁽¹⁺¹⁾ + 14a + a² - 6a + 14 + 1

= a³ - 6a² + 14a + a² - 6a + 14 + 1

= a³ + (- 6a² + a²) + (14a - 6a) + (14 + 1)

= a³ - 5a² + 8a + 15

= Dividend

Dividing 3x⁴ + 6x³ - 6x² + 2x - 7 by x - 3

$$\begin{array}{l} \phantom{x - 3)}{3x^3 + 15x^2 + 39x + 119} \\ x - 3\overline{\smash{\big)}3x^4 + 6x^3 - 6x^2 + 2x - 7} \\ \phantom{x - 3}\underline{\underset{-}{}3x^4 \underset{+}{-}9x^3} \\ \phantom{{x - 3)}{3x^4567}}15x^3 - 6x^2 + 2x - 7 \\ \phantom{{x - 3}{3x^456}}\underline{\underset{-}{+}15x^3 \underset{+}{-} 45x^2} \\ \phantom{{x - 3}{3x^456 + 15x^3 + }}39x^2 + 2x - 7 \\ \phantom{{x - 3}{3x^456 + 15x^3}}\underline{\underset{-}{+}39x^2 \underset{+}{-} 117x} \\ \phantom{{x - 3}{3x^456 + 15x^3 + 39x^2 +}}119x - 7 \\ \phantom{{x - 3}{3x^456 + 15x^3 + 39x^2}}\underline{\underset{-}{+}119x \underset{+}{-} 357} \\ \phantom{{x - 3}{3x^456 + 15x^3 + 39x^2 + 117x +}}350 \end{array}$$

Quotient = 3x³ + 15x² + 39x + 119

Remainder = 350

Verification:

Quotient x Divisor + Remainder

= (3x³ + 15x² + 39x + 119) $\times$ (x - 3) + 350

= x $\times$ (3x³ + 15x² + 39x + 119) - 3 $\times$ (3x³ + 15x² + 39x + 119) + 350

= 3x⁽³⁺¹⁾ + 15x⁽²⁺¹⁾ + 39x⁽¹⁺¹⁾ + 119x - 9x³ - 45x² - 117x - 357 + 350

= 3x⁴ + 15x³ + 39x² + 119x - 9x³ - 45x² - 117x - 357 + 350

= 3x⁴ + (15x³ - 9x³) + (39x² - 45x²) + (119x - 117x) + (- 357 + 350)

= 3x⁴ + 6x³ - 6x²+ 2x - 7

= Dividend

(6x² + x - 15) ÷ (3x + 5)

Dividing 6x² + x - 15 by 3x + 5

$$\begin{array}{l} \phantom{3x + 5)}{2x - 3} \\ 3x + 5\overline{\smash{\big)}6x^2 + x - 15} \\ \phantom{3x + 5}\underline{\underset{-}{}6x^2 \underset{-}{+}10x} \\ \phantom{{3x + 5}6x^2 +}-9x - 15 \\ \phantom{{3x + 5}6x^2 +}\underline{\underset{+}{-}9x \underset{+}{-} 15} \\ \phantom{{3x + 5}6x^2 + 9x -} \times \end{array}$$

Quotient = 2x - 3

Remainder = 0

Verification:

Quotient x Divisor + Remainder

= (2x - 3) $\times$ (3x + 5) + 0

= 2x $\times$ (3x + 5) - 3 $\times$ (3x + 5)

= 6x⁽¹⁺¹⁾ + 10x - 9x - 15

= 6x² + (10x - 9x) - 15

= 6x² + x - 15

= Dividend

Area of the rectangle = x³ - 8x² + 7

One side = (x - 1)

As we know the area of rectangle = One side x Other side

Other side = Area of rectangle ÷ One side

= (x³ - 8x² + 7) ÷ (x - 1)

$$\begin{array}{l} \phantom{x - 1)}{x^2 - 7x - 7} \\ x - 1\overline{\smash{\big)}x^3 - 8x^2 + 7} \\ \phantom{x - 1}\underline{\underset{-}{}x^3 \underset{+}{-}x^2} \\ \phantom{{x - 1}12x}-7x^2 + 7 \\ \phantom{{x - 1}221x}\underline{\underset{+}{-}7x^2 \phantom{+ 7} \underset{-}{+} 7x} \\ \phantom{{x - 1}{2x^3+}{+5x}}-7x + 7 \\ \phantom{{x - 1}{2x^3+}{+5x^2}}\underline{\underset{+}{-}7x \underset{-}{+} 7} \\ \phantom{{x - 1}{2x^3+}{+5x^2}{-7x}} \times \end{array}$$

Hence, length of adjacent side is x² - 7x - 7.

$x - \overline{x - y}$

= $x - x + y$

= $y$

Hence, option 3 is the correct option.

(5x - 4y) - (5y - 4x)

= 5x - 4y - 5y + 4x

= (5x + 4x) + (-4x - 5y)

= 9x + (- 9y)

= 9x - 9y

= 9(x - y)

Hence, option 1 is the correct option.

$x (y - x - \overline{x - y})$

= $x (y - x - x + y)$

= $x (2y - 2x)$

= $2x (y - x)$

Hence, option 2 is the correct option.

$(2x - y) + \overline{2x - y}$

= $2x - y + 2x - y$

= $4x - 2y$

Hence, option 4 is the correct option.

x(y - z) + y(z - x) - z(y - x)

= xy - xz + yz - xy - yz + xz

= (xy - xy) + (- xz + xz) + (yz - yz)

= 0 + 0 + 0

Hence, option 1 is the correct option.

a² - 2a + {5a² - (3a - 4a²)}

= a² - 2a + {5a² - 3a + 4a²}

= a² - 2a + {(5a² + 4a²) - 3a}

= a² - 2a + {9a² - 3a}

= a² - 2a + 9a² - 3a

= (a² + 9a²) + ( - 2a - 3a)

= 10a² + (- 5a)

= 10a² - 5a

Hence, a² - 2a + {5a² - (3a - 4a²)} = 10a² - 5a

$x - y - \lbrace x - y - (x + y) - \overline{x - y}\rbrace$

= $x - y - \lbrace x - y - (x + y) - x + y\rbrace$

= $x - y - \lbrace \cancel{x} - \cancel{y} - (x + y) - \cancel{x} + \cancel{y}\rbrace$

= $x - y - \lbrace -(x + y)\rbrace$

= $x - y - \lbrace - x - y\rbrace$

= $x - \cancel{y} + x + \cancel{y}$

= $2x$

Hence, $x - y - \lbrace x - y - (x + y) - \overline{x - y}\rbrace = 2x$

-3(1 - x²) - 2{x² - (3 - 2x²)}

= -3(1 - x²) - 2{x² - 3 + 2x²}

= -3(1 - x²) - 2{(x² + 2x²) - 3}

= -3(1 - x²) - 2{3x² - 3}

= -3 + 3x² - 6x² + 6

= (-3 + 6) + (3x² - 6x²)

= 3 + (-3x²)

= 3 - 3x²

Hence, -3(1 - x²) - 2{x² - (3 - 2x²)} = 3 - 3x²

$2\lbrace m - 3(n + \overline{m - 2n})\rbrace$

= $2\lbrace m - 3(n + m - 2n)\rbrace$

= $2\lbrace m - 3((n - 2n) + m)\rbrace$

= $2\lbrace m - 3(-n + m)\rbrace$

= $2\lbrace m + 3n - 3m\rbrace$

= $2\lbrace (m - 3m) + 3n\rbrace$

= $2\lbrace -2m + 3n\rbrace$

= $-4m + 6n$

Hence, $2\lbrace m - 3(n + \overline{m - 2n})\rbrace = 6n - 4m$

$3x - [3x - \lbrace 3x - (3x - \overline{3x - y})\rbrace]$

= $3x - [3x - \lbrace 3x - (3x - 3x + y)\rbrace]$

= $3x - [3x - \lbrace 3x - (y)\rbrace]$

= $3x - [3x - \lbrace 3x - y\rbrace]$

= $3x - [3x - 3x + y]$

= $3x - [y]$

= $3x - y$

Hence, $3x - [3x - \lbrace 3x - (3x - \overline{3x - y})\rbrace] = 3x - y$

$p^2x - 2\lbrace px - 3x(x^2 - \overline{3a - x^2})\rbrace$

= $p^2x - 2\lbrace px - 3x(x^2 - 3a + x^2)\rbrace$

= $p^2x - 2\lbrace px - 3x((x^2 + x^2) - 3a)\rbrace$

= $p^2x - 2\lbrace px - 3x(2x^2 - 3a)\rbrace$

= $p^2x - 2\lbrace px - 6x^3 + 9ax\rbrace$

= $p^2x - 2px + 12x^3 - 18ax$

Hence, $p^2x - 2\lbrace px - 3x(x^2 - \overline{3a - x^2})\rbrace = p^2x - 2px + 12x^3 - 18ax$

$2[6 + 4\lbrace m - 6(7 - \overline{n + p}) + q\rbrace]$

= $2[6 + 4\lbrace m - 6(7 - n - p) + q\rbrace]$

= $2[6 + 4\lbrace m - 42 + 6n + 6p + q\rbrace]$

= $2[6 + 4m - 168 + 24n + 24p + 4q]$

= $2[(6 - 168) + 4m + 24n + 24p + 4q]$

= $2[-162 + 4m + 24n + 24p + 4q]$

= $-324 + 8m + 48n + 48p + 8q$

Hence, $2[6 + 4\lbrace m - 6(7 - \overline{n + p}) + q\rbrace] = 8m + 48n + 48p + 8q -324$

$a - [a - \overline{b + a} - \lbrace a - (a - \overline{b - a})\rbrace]$

= $a - [a - \overline{b + a} - \lbrace a - (a - b + a)\rbrace]$

= $a - [a - \overline{b + a} - \lbrace a - (a + a - b)\rbrace]$

= $a - [a - \overline{b + a} - \lbrace a - (2a - b)\rbrace]$

= $a - [a - \overline{b + a} - \lbrace a - 2a + b\rbrace]$

= $a - [a - \overline{b + a} - \lbrace -a + b\rbrace]$

= $a - [a - \overline{b + a} + a - b]$

= $a - [a - b - a + a - b]$

= $a - [(a - a + a) + (- b - b)]$

= $a - [ a + (- 2b)]$

= $a - [ a - 2b]$

= $a - a + 2b$

= $2b$

Hence, $a - [a - \overline{b + a} - \lbrace a - (a - \overline{b - a})\rbrace] = 2b$

$3x - [4x - \overline{3x - 5y} - 3\lbrace 2x - (3x - \overline{2x - 3y})\rbrace]$

= $3x - [4x - \overline{3x - 5y} - 3\lbrace 2x - (3x - 2x + 3y)\rbrace]$

= $3x - [4x - \overline{3x - 5y} - 3\lbrace 2x - (x + 3y)\rbrace]$

= $3x - [4x - \overline{3x - 5y} - 3\lbrace 2x - x - 3y\rbrace]$

= $3x - [4x - \overline{3x - 5y} - 3\lbrace x - 3y\rbrace]$

= $3x - [4x - \overline{3x - 5y} - 3x + 9y]$

= $3x - [4x - 3x + 5y - 3x + 9y]$

= $3x - [(4x - 3x - 3x) + (5y + 9y)]$

= $3x - [-2x + 14y]$

= $3x + 2x - 14y$

= $5x - 14y$

Hence, $3x - [4x - \overline{3x - 5y} - 3\lbrace 2x - (3x - \overline{2x - 3y})\rbrace] = 5x - 14y$

a⁵ ÷ a³ + 3a x 2a

= a^(5-3) + 3a x 2a

= a² + 3a x 2a

= a² + 6a⁽¹⁺¹⁾

= a² + 6a²

= 7a²

Hence, a⁵ ÷ a³ + 3a x 2a = 7a²

$$\begin{aligned}x^5 ÷ (x^2 \times y^2) \times y^3\\[1em] = \frac{x^5}{x^2y^2} \times y^3\\[1em] = \frac{x^(5-2)}{y^2} \times y^3\\[1em] = \frac{x^3}{y^2} \times y^3\\[1em] = \frac{x^3y^3}{y^2}\\[1em] = x^3y^{3-2}\\[1em] = x^3y^1\\[1em]\end{aligned}$$

Hence, x⁵ ÷ (x² $\times$ y²) $\times$ y³ = x³y

(x⁵ ÷ x²) $\times$ y² $\times$ y³

= x^(5-2) $\times$ y² $\times$ y³

= x³ $\times$ y² $\times$ y³

= x³ $\times$ y⁽²⁺³⁾

= x³ $\times$ y⁵

Hence, (x⁵ ÷ x²) $\times$ y² $\times$ y³ = x³y⁵

(y³ - 5y²) ÷ y x (y - 1)

= (y³ ÷ y - 5y² ÷ y) x (y - 1)

= (y^(3-1) - 5y^(2-1)) x (y - 1)

= (y² - 5y¹) x (y - 1)

= y² x (y - 1) - 5y¹ x (y - 1)

= y⁽²⁺¹⁾ - y² - 5y⁽¹⁺¹⁾ - 5y

= y³ - y² - 5y² - 5y

= y³ - 6y² - 5y

Hence, (y³ - 5y²) ÷ y x (y - 1) = y³ - 6y² - 5y

(-18xy) - (-8xy)

= -18xy + 8xy

= - 10xy

Hence, option 2 is the correct option.

(9a + 7b - 6c) - (2a - 3b + 4c)

= 9a + 7b - 6c - 2a + 3b - 4c

= (9a - 2a) + (7b + 3b) + (- 6c - 4c)

= 7a + 10b + (-10c)

= 7a + 10b -10c

Hence, option 2 is the correct option.

-81a⁵b⁴c³ ÷ (-9a²b²c)

= $\frac{-81}{-9}$ a^(5-2)b^(4-2)c^(3-1)

= 9 a³b²c²

Hence, option 4 is the correct option.

$-(-pq - \overline{p^2 - pq})$

= $-(-pq - p^2 + pq)$

= $-((-pq + pq) - p^2)$

= $-(- p^2)$

= $p^2$

Hence, option 3 is the correct option.

$x^3 - y^3 - x(y^2 + \overline{x^2 - z^2})$

= $x^3 - y^3 - x(y^2 + x^2 - z^2)$

= $x^3 - y^3 - xy^2 - x^3 + xz^2$

= $(x^3 - x^3) - y^3 - xy^2 + xz^2$

= $- y^3 - xy^2 + xz^2$

Hence, option 4 is the correct option.

Given :

2x⁴ - 3x² + $\frac{7}{\text{x}}$

A constant term is a term that does not contain the variable in this case, x.

Thus, the expression contains no constant term.

So, statement 1 is true.

In an algebraic expression in terms of one variable, the term(s) independent of the variable is called the constant.

This is a standard definition in algebra: a constant term is the part of the expression that does not change with the variable i.e., it is independent of the variable.

So, statement 2 is true.

Hence, option 1 is the correct option.

An algebraic expression which contains three different terms is called a trinomial.

So, reason (R) is true.

Given, 5x + y² - x³, xy + yz + zx, x² - x + 1

All three expressions have three distinct terms.

So, assertion (A) is true and, reason (R) is the correct explanation of assertion (A).

Hence, option 1 is the correct option.

An algebraic expression is a polynomial if the degree of each term used in it is a whole number.

So, reason (R) is true

A polynomial in one variable (say, x) must have:

• Only non-negative integers (i.e., whole numbers) as exponents of x.

• No negative or fractional powers.

So, reason (R) is true.

Given,

Algebraic expressions : 3x + 9, 9 - $\frac{3}{\text{x}^2}$, 100

3x + 9 and 100 (constant polynomial) both are polynomials, while 9 - $\frac{3}{\text{x}^2}$ is not as on taking the variable in numerator (9 - 3x^-2) the power becomes negative.

So, assertion (A) is false.

Hence, option 4 is the correct option.

An algebraic expression having two or more variables, the highest sum of the powers of all the variables in each term taken as the degree of the polynomial if this sum is a whole number.

The maximum of these sums gives the degree. These powers must be whole numbers for it to be a valid polynomial.

So, reason (R) is true.

Given; 2xyz + 3x²y

Terms; 2xyz, 3x²y

For term 2xyz, degree = 1 + 1 + 1 = 3

For term 3x²y, degree = 2 + 1 = 3

The degree of a polynomial in multiple variables is the highest total degree among its terms.

Thus, the degree of this polynomial is 3.

Therefore, this is a cubic polynomial in three variables.

So, assertion (A) is true and, reason (R) is the correct explanation of assertion (A).

Hence, option 1 is the correct option.

A trinomial is an algebraic expression with exactly three terms.

Here, 2024x²yz is just a single term — a monomial.

Thus, the statement that it's not a trinomial is correct.

So, assertion (A) is true.

For combining polynomials, the like terms of the given polynomials are combined together.

Like terms are terms with the same variables raised to the same powers.

When combining polynomials, you add or subtract like terms.

So, reason (R) is true but, reason (R) does not explains assertion(A).

Hence, option 2 is the correct option.

(i) Coefficient = $\frac{1}{7}$ xy²z³

(ii) Coefficient = xy²z³

(iii) Coefficient = $\frac{1}{7}$ y²z³

(iv) Coefficient = $\frac{5}{7}$ z³

(v) Coefficient = $\frac{5}{7}$ xy²

(vi) Coefficient = $\frac{5}{7}$ y²

(vii) Coefficient = $\frac{1}{7}$ z³

(viii) Coefficient = 5 xyz²

(ix) Coefficient = $\frac{5}{7}$ xy²z²

(x) Coefficient = $\frac{5}{7}$ xyz

(xi) Coefficient = $\frac{1}{7}$ yz²

Like terms are those terms which have the same variable raised to the same power.

Hence, 3xy,- 8yx ; 2y²x, - 3.2y²x ; -4x²y, 2.5x²y and x²y are like terms.

Like terms are those terms which have the same variable raised to the same power.

Hence, y²z³, - 4y²z³ and 2z³y²; xy²z³, - 8xz³y² ; - 5x²yz, 3x²yz are like terms.

The sides of a triangle = (x² - 3xy + 8), (4x² + 5xy - 3) and (6 - 3x² + 4xy)

Perimeter of triangle = sum of all its side.

(x² - 3xy + 8) + (4x² + 5xy - 3) + (6 - 3x² + 4xy)

= x² - 3xy + 8 + 4x² + 5xy - 3 + 6 - 3x² + 4xy

= (x² + 4x² - 3x²) + (- 3xy + 5xy + 4xy) + (- 3 + 6 + 8)

= 2x² + 6xy + 11

Hence, perimeter of triangle = 2x² + 6xy + 11

The perimeter of a triangle = 8y² - 9y + 4

Two sides are 3y² - 5y and 4y² + 12

Let the third side be S.

As we know that, perimeter of triangle = sum of all its side.

8y² - 9y + 4 = (3y² - 5y) + (4y² + 12) + S

⇒ 8y² - 9y + 4 = 3y² - 5y + 4y² + 12 + S

⇒ 8y² - 9y + 4 = (3y² + 4y²) - 5y + 12 + S

⇒ 8y² - 9y + 4 = 7y² - 5y + 12 + S

⇒ S = (8y² - 9y + 4) - (7y² - 5y + 12)

⇒ S = 8y² - 9y + 4 - 7y² + 5y - 12

⇒ S = (8y² - 7y²) + (- 9y + 5y) + (4 - 12)

⇒ S = 1y² - 4y - 8

Hence, the third side of triangle = (y² - 4y - 8)

The two adjacent sides of a rectangle = (2x² - 5xy + 3z²) and (4xy - x² - z²)

As we know the perimeter of rectangle = 2 x (l + b)

= 2 $\times$ [(2x² - 5xy + 3z²) + (4xy - x² - z²)]

= 2 $\times$ [2x² - 5xy + 3z² + 4xy - x² - z²]

= 2 $\times$ [(2x² - x²) + (- 5xy + 4xy) + (3z² - z²)]

= 2 $\times$ [x² - 1xy + 2z²]

= 2 $\times$ x² - 2 $\times$ 1xy + 2 $\times$ 2z²

= 2x² - 2xy + 4z²

Hence, the perimeter of rectangle = 2x² - 2xy + 4z²

Let A be the number that should be subtracted from 19x⁴ + 2x³ + 30x - 37 to get 8x⁴ + 22x³ - 7x - 60

(19x⁴ + 2x³ + 30x - 37) - A = (8x⁴ + 22x³ - 7x - 60)

⇒ (19x⁴ + 2x³ + 30x - 37) - (8x⁴ + 22x³ - 7x - 60) = A

⇒ A = 19x⁴ + 2x³ + 30x - 37 - 8x⁴ - 22x³ + 7x + 60

⇒ A = (19x⁴ - 8x⁴) + (2x³ - 22x³) + (30x + 7x) + (- 37 + 60)

⇒ A = 11x⁴ - 20x³ + 37x + 23

Hence, the number is 11x⁴ - 20x³ + 37x + 23.

(22x - 20y -13z + 26) - (15x - 18y + 19z)

= 22x - 20y -13z + 26 - 15x + 18y - 19z

= (22x - 15x) + (- 20y + 18y) + (-13z - 19z) + 26

= 7x - 2y -32z + 26

Hence, (22x - 20y -13z +26) is less than (15x - 18y + 19z) by (7x - 2y -32z + 26)

(5x²y² - 18xy² - 10x²y) - (- 5x² + 6x²y - 7xy)

= 5x²y² - 18xy² - 10x²y + 5x² - 6x²y + 7xy

= 5x²y² - 18xy² + (- 10x²y - 6x²y) + 5x² + 7xy

= 5x²y² - 18xy² - 16x²y + 5x² + 7xy

Hence, (5x²y² - 18xy² - 10x²y) is bigger than (- 5x² + 6x²y - 7xy) by (5x²y² - 18xy² - 16x²y + 5x² + 7xy)

(-4x²y³) $\times$ (-5x²y⁵)

= (-4) $\times$ (-5)x⁽²⁺²⁾y⁽³⁺⁵⁾

= 20 x⁴y⁸

When x = 2 and y = 1,

= 20 x 2⁴ x 1⁸

= 20 x 16 x 1

= 320

Hence, if x = 2 and y = 1 ; the value of (-4x²y³) x (-5x²y⁵) = 320.

(3x - 2) (x + 5)

= 3x (x + 5) - 2 (x + 5)

= 3x⁽¹⁺¹⁾ + 15x - 2x - 10

= 3x² + (15x - 2x) - 10

= 3x² + 13x - 10

For x = 2

3 x 2² + 13 x 2 - 10

= 3 x 4 + 26 - 10

= 12 + 26 - 10

= 38 - 10

= 28

Hence, for x = 2, the value of (3x - 2) (x + 5) is 28.

(2x - 5y) (2x + 3y)

= 2x (2x + 3y) - 5y (2x + 3y)

= 4x⁽¹⁺¹⁾ + 6xy - 10xy - 15y⁽¹⁺¹⁾

= 4x² + (6xy - 10xy) - 15y²

= 4x² - 4xy - 15y²

For x = 2, y = 3

4 x 2² - 4 x 2 x 3 - 15 x 3²

= 4 x 4 - 4 x 2 x 3 - 15 x 9

= 16 - 24 - 135

= - 143

Hence, for x = 2, y = 3, the value (2x - 5y) (2x + 3y) = - 143.

xz (x² + y²)

= x⁽²⁺¹⁾z + xzy²

= x³z + xzy²

For x = 2, y = 1 and z = 1

2³ x 1 + 2 x 1 x 1²

= 8 x 1 + 2 x 1 x 1

= 8 + 2

= 10

Hence, for x = 2, y = 1 and z = 1, the value of xz (x² + y²) is 10.

x(x - 5) + 2

= x² - 5x + 2

For x = 1

= 1² - 5 x 1 + 2

= 1 - 5 + 2

= - 2

Hence, for x = 1, the value of x(x - 5) + 2 is - 2.

xy²(x - 5y) + 1

= x⁽¹⁺¹⁾y² - 5xy⁽²⁺¹⁾ + 1

= x²y² - 5xy³ + 1

For x = 2 and y = 1,

2² x 1² - 5 x 2 x 1³ + 1

= 4 x 1 - 5 x 2 x 1 + 1

= 4 - 10 + 1

= - 5

Hence, for x = 2 and y = 1, the value of [xy²(x - 5y) + 1] is - 5.

2x(3x - 5) - 5(x - 2) - 18

= 6x⁽¹⁺¹⁾ - 10x - 5x + 10 - 18

= 6x² + (- 10x - 5x) + (10 - 18)

= 6x² - 15x - 8

For x = 2,

= 6 x 2² - 15 x 2 - 8

= 6 x 4 - 15 x 2 - 8

= 24 - 30 - 8

= - 14

Hence, for x = 2, the value of [2x(3x - 5) - 5(x - 2) - 18] is - 14.

(-3x²y²) $\times$ (x - 2y)

= (-3x²y²) $\times$ x - (-3x²y²) $\times$ 2y

= (-3x⁽²⁺¹⁾y²) - (-6x²y⁽²⁺¹⁾)

= -3x³y² + 6x²y³

For x = 1 and y = 2

(-3x²y²) $\times$ (x - 2y) = -3x³y² + 6x²y³

Taking LHS:

(-3x²y²) $\times$ (x - 2y)

= (-3 x 1² x 2²) x (1 - 2 x 2)

= (-3 x 1 x 4) x (1 - 4)

= (-12) x (- 3)

= 36

Now, taking RHS:

-3x³y² + 6x²y³

= -3 x 1³ x 2² + 6 x 1² x 2³

= -3 x 1 x 4 + 6 x 1 x 8

= -12 + 48

= 36

Hence, LHS = RHS.

(2x² - 4x + 5) $\times$ (x² + 3x - 7)

= 2x² $\times$ (x² + 3x - 7) - 4x $\times$ (x² + 3x - 7) + 5 $\times$ (x² + 3x - 7)

= 2x⁽²⁺²⁾ + 6x⁽²⁺¹⁾ - 14x² - 4x⁽¹⁺²⁾ - 12x⁽¹⁺¹⁾ + 28x + 5x² + 15x - 35

= 2x⁴ + 6x³ - 14x² - 4x³ - 12x² + 28x + 5x² + 15x - 35

= 2x⁴ + (6x³ - 4x³) + (- 14x² - 12x² + 5x²) + (28x + 15x) - 35

= 2x⁴ + 2x³ - 21x² + 43x - 35

Hence, (2x² - 4x + 5) $\times$ (x² + 3x - 7) = 2x⁴ + 2x³ - 21x² + 43x - 35

(ab - 1) (3 - 2ab)

= ab (3 - 2ab) - 1 (3 - 2ab)

= 3ab - 2a⁽¹⁺¹⁾b⁽¹⁺¹⁾ - 3 + 2ab

= 3ab - 2a²b² - 3 + 2ab

= (3ab + 2ab) - 2a²b² - 3

= 5ab - 2a²b² - 3

Hence, (ab - 1) (3 - 2ab) = 5ab - 2a²b² - 3

(5 - x) (6 - 5x) (2 - x)

= [5 (6 - 5x) - x (6 - 5x)](2 - x)

= [30 - 25x - 6x + 5x⁽¹⁺¹⁾](2 - x)

= [30 + (- 25x - 6x) + 5x²](2 - x)

= [30 - 31x + 5x²](2 - x)

= 2 (30 - 31x + 5x²) - x (30 - 31x + 5x²)

= 60 - 62x + 10x² - 30x + 31x⁽¹⁺¹⁾ - 5x⁽²⁺¹⁾

= 60 - 62x + 10x² - 30x + 31x² - 5x³

= 60 + (- 62x - 30x) + (10x² + 31x²) - 5x³

= 60 - 92x + 41x² - 5x³

Hence, (5 - x) (6 - 5x) (2 - x) = 60 - 92x + 41x² - 5x³

The product = (16x⁴ -1)

One number = (2x - 1)

Let the other number be A.

(2x - 1) $\times$ A = (16x⁴ -1)

⇒ A = $\frac{(16x^4 - 1)}{(2x - 1)}$

⇒ A = $\frac{(4x^2 - 1)(4x^2 + 1)}{(2x - 1)}$

⇒ A = $\frac{(2x - 1)(2x + 1)(4x^2 + 1)}{(2 - 1)}$

⇒ A = $\frac{\cancel{(2x - 1)}(2x + 1)(4x^2 + 1)}{\cancel{(2x - 1)}}$

⇒ A = (2x + 1)(4x² + 1)

⇒ A = 2x(4x² + 1) + 1(4x² + 1)

⇒ A = 8x⁽²⁺¹⁾ + 2x + 4x² + 1

⇒ A = 8x³ + 4x² + 2x + 1

Hence, the other number is 8x³ + 4x² + 2x + 1.

The product of (x² + xy + y²) and (x - y)

= (x² + xy + y²) $\times$ (x - y)

= x $\times$ (x² + xy + y²) - y $\times$ (x² + xy + y²)

= x⁽¹⁺²⁾ + x⁽¹⁺¹⁾y + xy² - x²y - xy⁽¹⁺¹⁾ - y⁽¹⁺²⁾

= x³ + x²y + xy² - x²y - xy² - y³

= x³ + (x²y - x²y) + (xy² - xy²) - y³

= x³ - y³

Now, (x⁶ - y⁶) ÷ ( x³ - y³)

$$\begin{array}{l} \phantom{x^3 - y^3)}{x^3 + y^3} \\ x^3 - y^3\overline{\smash{\big)}x^6 - y^6 \phantom{+ x^3y^3}} \\ \phantom{x^3 - y^3}\underline{\underset{-}{}x^6 \phantom{- y^6} \underset{+}{-}x^3y^3} \\ \phantom{{x^3 - y^3}6x^2 +}x^3y^3 - y^6 \\ \phantom{{x^3 - y^3}6x^21}\underline{\underset{-}{+}x^3y^3 \underset{+}{-} y^6} \\ \phantom{{x^3 - y^3}6x^2 + 9x -} \times \end{array}$$

Hence, the answer is (x³ + y³)

$3a \times [8b ÷ 4 - 6\lbrace a - (5a - \overline{3b - 2a})\rbrace]$

= $3a \times [8b ÷ 4 - 6\lbrace a - (5a - 3b + 2a)\rbrace]$

= $3a \times [8b ÷ 4 - 6\lbrace a - (5a + 2a - 3b )\rbrace]$

= $3a \times [8b ÷ 4 - 6\lbrace a - (7a - 3b )\rbrace]$

= $3a \times [8b ÷ 4 - 6\lbrace a - 7a + 3b \rbrace]$

= $3a \times [8b ÷ 4 - 6\lbrace -6a + 3b \rbrace]$

= $3a \times [8b ÷ 4 + 36a - 18b ]$

= $3a \times [2b + 36a - 18b]$

= $3a \times [2b - 18b + 36a]$

= $3a \times [-16b + 36a]$

= $-48ab + 108a^2$

Hence, $3a \times [8b ÷ 4 - 6\lbrace a - (5a - \overline{3b - 2a})\rbrace] = 108a^2 - 48ab$

7x + 4{x² ÷ (5x ÷ 10)} - 3{2 - x³ ÷ (3x² ÷ x)}

= 7x + 4{x² ÷ (5x ÷ 10)} - 3{2 - x³ ÷ 3x}

= 7x + 4{x² ÷ (5x ÷ 10)} - 3{2 - $\frac{\text{x}^2}{3}$}

= 7x + 4{x² ÷ (5x ÷ 10)} - 6 + $\frac{3\text{x}^2}{3}$

= 7x + 4{x² ÷ $\frac{5\text{x}}{10}$} - 6 + x²

= 7x + 4{x² ÷ $\frac{\text{x}}{2}$} - 6 + x²

= 7x + 4{$\frac{2\text{x}^2}{\text{x}}$} - 6 + x²

= 7x + 4 x 2x - 6 + x²

= 7x + 8x - 6 + x²

= x² + 15x - 6

Hence, the answer is x² + 15x - 6