Factorization

HCF of $6xy$ and $4xz$ = (H.C.F. of 6 and 4) × (H.C.F. of $xy$ and $xz$)

= 2 × $x$

= $2x$

So, $2x$ is a common factor of each term.

Factorizing,

$$\begin{aligned}\Rightarrow 6xy - 4xz \\[1em] \Rightarrow 2x(3y - 2z)\end{aligned}$$

Hence, $\bm{6xy - 4xz = 2x(3y - 2z)}$.

HCF of $9bx^2$ and $15b^2x$ = (H.C.F. of 9 and 15) × (H.C.F. of $bx^2$ and $b^2x$)

= 3 × $bx$

= $3bx$

So, $3bx$ is a common factor of each term.

Factorizing,

$$\begin{aligned}\Rightarrow 9bx^2 + 15b^2x \\[1em] \Rightarrow 3bx(3x + 5b)\end{aligned}$$

Hence, $\bm{9bx^2 + 15b^2x = 3bx(3x + 5b)}$.

HCF of $4a^2b, 6ab^2$ and $8ab$ = (H.C.F. of 4, 6 and 8) × (H.C.F. of $a^2b, ab^2$ and $ab$)

= 2 × $ab$

= $2ab$

So, $2ab$ is a common factor of each term.

Factorizing,

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

Hence, $\bm{4a^2b - 6ab^2 + 8ab = 2ab(2a - 3b + 4)}$.

HCF of $5x^2, 15x^3$ and $20x$ = (H.C.F. of 5, 15 and 20) × (H.C.F. of $x^2, x^3$ and $x$)

= 5 × $x$

= $5x$

So, $5x$ is a common factor of each term.

Factorizing,

$$\begin{aligned}\Rightarrow 5x^2 + 15x^3 - 20x \\[1em] \Rightarrow 5x(x + 3x^2 - 4)\end{aligned}$$

Hence, $\bm{5x^2 + 15x^3 - 20x = 5x(x + 3x^2 - 4)}$.

HCF of $2x^3, x^2$ and $4x$ = (H.C.F. of 2, 1 and 4) × (H.C.F. of $x^3, x^2$ and $x$)

= 1 × $x$

= $x$

So, $x$ is a common factor of each term.

Factorizing,

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

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

HCF of $a^3b, a^2b^2$ and $b^3$ = $b$

So, $b$ is a common factor of each term.

Factorizing,

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

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

HCF of $28a^2b^2c$ and $42ab^2c^2$ = (H.C.F. of 28 and 42) × (H.C.F. of $a^2b^2c$ and $ab^2c^2$)

= 14 × $ab^2c$

= $14ab^2c$

So, $14ab^2c$ is a common factor of each term.

Factorizing,

$$\begin{aligned}\Rightarrow 28a^2b^2c - 42ab^2c^2 \\[1em] \Rightarrow 14ab^2c(2a - 3c)\end{aligned}$$

Hence, $\bm{28a^2b^2c - 42ab^2c^2 = 14ab^2c(2a - 3c)}$.

HCF of $15x^2y, 6xy^2$ and $9y^3$ = (H.C.F. of 15, 6 and 9)×(H.C.F. of $x^2y, xy^2$ and $y^3$)

= 3 × $y$

= $3y$

So, $3y$ is a common factor of each term.

Factorizing,

$$\begin{aligned}\Rightarrow 15x^2y - 6xy^2 + 9y^3 \\[1em] \Rightarrow 3y(5x^2 - 2xy + 3y^2)\end{aligned}$$

Hence, $\bm{15x^2y - 6xy^2 + 9y^3 = 3y(5x^2 - 2xy + 3y^2)}$.

HCF of $10x^2y, 15xy^2$ and $25x^2y^2$ = (H.C.F. of 10, 15 and 25) × (H.C.F. of $x^2y, xy^2$ and $x^2y^2$)

= 5 × $xy$

= $5xy$

So, $5xy$ is a common factor of each term.

Factorizing,

$$\begin{aligned}\Rightarrow 10x^2y + 15xy^2 - 25x^2y^2 \\[1em] \Rightarrow 5xy(2x + 3y - 5xy)\end{aligned}$$

Hence, $\bm{10x^2y + 15xy^2 - 25x^2y^2 = 5xy(2x + 3y - 5xy)}$.

HCF of $18x^2y$ and $24xyz$ = (H.C.F. of 18 and 24) × (H.C.F. of $x^2y$ and $xyz$)

= 6 × $xy$

= $6xy$

So, $6xy$ is a common factor of each term.

Factorizing,

$$\begin{aligned}\Rightarrow 18x^2y - 24xyz \\[1em] \Rightarrow 6xy(3x - 4z)\end{aligned}$$

Hence, $\bm{18x^2y - 24xyz = 6xy(3x - 4z)}$.

HCF of $27a^3b^3$ and $45a^4b^2$ = (H.C.F. of 27 and 45) × (H.C.F. of $a^3b^3$ and $a^4b^2$)

= 9 × $a^3b^2$

= $9a^3b^2$

So, $9a^3b^2$ is a common factor of each term.

Factorizing,

$$\begin{aligned}\Rightarrow 27a^3b^3 - 45a^4b^2 \\[1em] \Rightarrow 9a^3b^2(3b - 5a)\end{aligned}$$

Hence, $\bm{27a^3b^3 - 45a^4b^2 = 9a^3b^2(3b - 5a)}$.

HCF of $4(a + b)$ and $6(a + b)^2$ = (H.C.F. of 4 and 6) × (H.C.F. of $(a + b)$ and $(a + b)^2$)

= 2 × $(a + b)$

= $2(a + b)$

So, $2(a + b)$ is a common factor of each term.

Factorizing,

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

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

HCF of $2a(3x + 5y)$ and $5b(3x + 5y)$ = (H.C.F. of 2 and 5) × (H.C.F. of $a(3x + 5y)$ and $b(3x + 5y)$)

= 1 × $(3x + 5y)$

= $(3x + 5y)$

So, $(3x + 5y)$ is a common factor of each term.

Factorizing,

$$\begin{aligned}\Rightarrow 2a(3x + 5y) - 5b(3x + 5y) \\[1em] \Rightarrow (3x + 5y)(2a - 5b)\end{aligned}$$

Hence, $\bm{2a(3x + 5y) - 5b(3x + 5y) = (3x + 5y)(2a - 5b)}$.

HCF of $2x(p^2 + q^2)$ and $4y(p^2 + q^2)$ = (H.C.F. of 2 and 4) × (H.C.F. of $x(p^2 + q^2)$ and $y(p^2 + q^2)$)

= 2 × $(p^2 + q^2)$

= $2(p^2 + q^2)$

So, $2(p^2 + q^2)$ is a common factor of each term.

Factorizing,

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

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

HCF of $8(3a - 2b)^2$ and $10(3a - 2b)$ = (H.C.F. of 8 and 10) × (H.C.F. of $(3a - 2b)^2$ and $(3a - 2b)$)

= 2 × $(3a - 2b)$

= $2(3a - 2b)$

So, $2(3a - 2b)$ is a common factor of each term.

Factorizing,

$$\begin{aligned}\Rightarrow 8(3a - 2b)^2 - 10(3a - 2b) \\[1em] \Rightarrow 2(3a - 2b)[4(3a - 2b) - 5] \\[1em] \Rightarrow 2(3a - 2b)(12a - 8b - 5)\end{aligned}$$

Hence, $\bm{8(3a - 2b)^2 - 10(3a - 2b) = 2(3a - 2b)(12a - 8b - 5)}$.

HCF of $x(x + y)^3$ and $3x^2y(x + y)$ = (H.C.F. of 1 and 3) × (H.C.F. of $x(x + y)^3$ and $x^2y(x + y)$)

= 1 × $x(x + y)$

= $x(x + y)$

So, $x(x + y)$ is a common factor of each term.

Factorizing,

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

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

As $(2y - 3x) = -(3x - 2y)$, the given expression can be written as $a(3x - 2y) - b(3x - 2y)$.

HCF of $a(3x - 2y)$ and $b(3x - 2y)$ = $(3x - 2y)$

So, $(3x - 2y)$ is a common factor of each term.

Factorizing,

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

Hence, $\bm{a(3x - 2y) + b(2y - 3x) = (3x - 2y)(a - b)}$.

As $(c - b) = -(b - c)$, the given expression can be written as $a(b - c)^2 + d(b - c)$.

HCF of $a(b - c)^2$ and $d(b - c)$ = $(b - c)$

So, $(b - c)$ is a common factor of each term.

Factorizing,

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

Hence, $\bm{a(b - c)^2 - d(c - b) = (b - c)(ab - ac + d)}$.

HCF of $a(4a - b), 2b(4a - b)$ and $c(4a - b)$ = (H.C.F. of 1, 2 and 1) × (H.C.F. of $a(4a - b), b(4a - b)$ and $c(4a - b)$)

= 1 × $(4a - b)$

= $(4a - b)$

So, $(4a - b)$ is a common factor of each term.

Factorizing,

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

Hence, $\bm{a(4a - b) + 2b(4a - b) - c(4a - b) = (4a - b)(a + 2b - c)}$.

HCF of $m(mx + ny)^2, mn(mx + ny)$ and $n(mx + ny)$ = $(mx + ny)$

So, $(mx + ny)$ is a common factor of each term.

Factorizing,

$$\begin{aligned}\Rightarrow m(mx + ny)^2 + mn(mx + ny) + n(mx + ny) \\[1em] \Rightarrow (mx + ny)[m(mx + ny) + mn + n] \\[1em] \Rightarrow (mx + ny)(m^2x + mny + mn + n)\end{aligned}$$

Hence, $\bm{m(mx + ny)^2 + mn(mx + ny) + n(mx + ny) =}$
$\bm{(mx + ny)(m^2x + mny + mn + n)}$.

Factorizing,

$$\begin{aligned}\Rightarrow 5xy + 20x - 9y - 36 \\[1em] \Rightarrow (5xy + 20x) - (9y + 36) \\[1em] \Rightarrow 5x(y + 4) - 9(y + 4) \\[1em] \Rightarrow (y + 4)(5x - 9)\end{aligned}$$

Hence, $\bm{5xy + 20x - 9y - 36 = (y + 4)(5x - 9)}$.

Factorizing,

$$\begin{aligned}\Rightarrow 5ab - 10b - 7a + 14 \\[1em] \Rightarrow (5ab - 10b) - (7a - 14) \\[1em] \Rightarrow 5b(a - 2) - 7(a - 2) \\[1em] \Rightarrow (a - 2)(5b - 7)\end{aligned}$$

Hence, $\bm{5ab - 10b - 7a + 14 = (a - 2)(5b - 7)}$.

Factorizing,

$$\begin{aligned}\Rightarrow 9xy + 7y - 9y^2 - 7x \\[1em] \Rightarrow (9xy - 9y^2) + (7y - 7x) \\[1em] \Rightarrow 9y(x - y) - 7(x - y) \\[1em] \Rightarrow (x - y)(9y - 7)\end{aligned}$$

Hence, $\bm{9xy + 7y - 9y^2 - 7x = (x - y)(9y - 7)}$.

Factorizing,

$$\begin{aligned}\Rightarrow 3ax - 6ay - 8by + 4bx \\[1em] \Rightarrow (3ax - 6ay) + (4bx - 8by) \\[1em] \Rightarrow 3a(x - 2y) + 4b(x - 2y) \\[1em] \Rightarrow (x - 2y)(3a + 4b)\end{aligned}$$

Hence, $\bm{3ax - 6ay - 8by + 4bx = (x - 2y)(3a + 4b)}$.

Factorizing,

$$\begin{aligned}\Rightarrow 4x^2 - 10xy - 6xz + 15yz \\[1em] \Rightarrow (4x^2 - 10xy) - (6xz - 15yz) \\[1em] \Rightarrow 2x(2x - 5y) - 3z(2x - 5y) \\[1em] \Rightarrow (2x - 5y)(2x - 3z)\end{aligned}$$

Hence, $\bm{4x^2 - 10xy - 6xz + 15yz = (2x - 5y)(2x - 3z)}$.

Factorizing,

$$\begin{aligned}\Rightarrow 3ab - 6bc - 3ax + 6cx \\[1em] \Rightarrow 3(ab - 2bc - ax + 2cx) \\[1em] \Rightarrow 3[(ab - 2bc) - (ax - 2cx)] \\[1em] \Rightarrow 3[b(a - 2c) - x(a - 2c)] \\[1em] \Rightarrow 3(a - 2c)(b - x)\end{aligned}$$

Hence, $\bm{3ab - 6bc - 3ax + 6cx = 3(b - x)(a - 2c)}$.

Factorizing,

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

Hence, $\bm{a^2 + ab(1 + b) + b^3 = (a + b)(a + b^2)}$.

Factorizing,

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

Hence, $\bm{xy^2 + (x - 1)y - 1 = (y + 1)(xy - 1)}$.

Factorizing,

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

Hence, $\bm{b^2 - ab(1 - a) - a^3 = (b - a)(b + a^2)}$.

Factorizing,

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

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

Factorizing,

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

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

Factorizing,

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

Hence, $\bm{x - 1 - (x - 1)^2 + 4 - 4x = -(x - 1)(x + 2)}$.

Factorizing,

$$\begin{aligned}\Rightarrow b(c - d)^2 - a(d - c) + 5c - 5d \\[1em] \Rightarrow b(c - d)^2 + a(c - d) + 5(c - d) \\[1em] \Rightarrow (c - d)[b(c - d) + a + 5] \\[1em] \Rightarrow (c - d)(bc - bd + a + 5)\end{aligned}$$

Hence, $\bm{b(c - d)^2 - a(d - c) + 5c - 5d = (c - d)(bc - bd + a + 5)}$.

Factorizing,

$$\begin{aligned}\Rightarrow ab(c^2 + d^2) - a^2cd - b^2cd \\[1em] \Rightarrow abc^2 + abd^2 - a^2cd - b^2cd \\[1em] \Rightarrow (abc^2 - a^2cd) - (b^2cd - abd^2) \\[1em] \Rightarrow ac(bc - ad) - bd(bc - ad) \\[1em] \Rightarrow (bc - ad)(ac - bd)\end{aligned}$$

Hence, $\bm{ab(c^2 + d^2) - a^2cd - b^2cd = (ac - bd)(bc - ad)}$.

Factorizing,

$$\begin{aligned}\Rightarrow ab(x^2 + y^2) + xy(a^2 + b^2) \\[1em] \Rightarrow abx^2 + aby^2 + a^2xy + b^2xy \\[1em] \Rightarrow (abx^2 + a^2xy) + (aby^2 + b^2xy) \\[1em] \Rightarrow ax(bx + ay) + by(ay + bx) \\[1em] \Rightarrow (bx + ay)(ax + by)\end{aligned}$$

Hence, $\bm{ab(x^2 + y^2) + xy(a^2 + b^2) = (bx + ay)(ax + by)}$.

Factorizing,

$$\begin{aligned}\Rightarrow ab(x^2 + y^2) - xy(a^2 + b^2) \\[1em] \Rightarrow abx^2 + aby^2 - a^2xy - b^2xy \\[1em] \Rightarrow (abx^2 - a^2xy) + (aby^2 - b^2xy) \\[1em] \Rightarrow ax(bx - ay) + by(ay - bx) \\[1em] \Rightarrow ax(bx - ay) - by(bx - ay) \\[1em] \Rightarrow (bx - ay)(ax - by)\end{aligned}$$

Hence, $\bm{ab(x^2 + y^2) - xy(a^2 + b^2) = (bx - ay)(ax - by)}$.

Using the identity,

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

And

[$(x - y)^2 = x^2 - 2xy + y^2$]

Factorizing,

$$\begin{aligned}\Rightarrow (ax + by)^2 + (bx - ay)^2 \\[1em] \Rightarrow (a^2x^2 + b^2y^2 + 2abxy) + (b^2x^2 + a^2y^2 - 2abxy) \\[1em] \Rightarrow a^2x^2 + b^2x^2 + b^2y^2 + a^2y^2 \\[1em] \Rightarrow (a^2x^2 + b^2x^2) + (a^2y^2 + b^2y^2) \\[1em] \Rightarrow x^2(a^2 + b^2) + y^2(a^2 + b^2) \\[1em] \Rightarrow (x^2 + y^2)(a^2 + b^2)\end{aligned}$$

Hence, $\bm{(ax + by)^2 + (bx - ay)^2 = (x^2 + y^2)(a^2 + b^2)}$.

Factorizing,

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

Hence, $\bm{a(a + b - c) - bc = (a - c)(a + b)}$.

Factorizing,

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

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

Factorizing,

$$\begin{aligned}\Rightarrow 3x^5 - 6x^4 - 2x^3 + 4x^2 + x - 2 \\[1em] \Rightarrow (3x^5 - 6x^4) - (2x^3 - 4x^2) + (x - 2) \\[1em] \Rightarrow 3x^4(x - 2) - 2x^2(x - 2) + 1(x - 2) \\[1em] \Rightarrow (x - 2)(3x^4 - 2x^2 + 1)\end{aligned}$$

Hence, $\bm{3x^5 - 6x^4 - 2x^3 + 4x^2 + x - 2 = (x - 2)(3x^4 - 2x^2 + 1)}$.

Factorizing,

$$\begin{aligned}\Rightarrow 2a + 2b - 3cx - 3bx + 2c - 3ax \\[1em] \Rightarrow (2a - 3ax) + (2b - 3bx) + (2c - 3cx) \\[1em] \Rightarrow a(2 - 3x) + b(2 - 3x) + c(2 - 3x) \\[1em] \Rightarrow (2 - 3x)(a + b + c)\end{aligned}$$

Hence, $\bm{2a + 2b - 3cx - 3bx + 2c - 3ax = (a + b + c)(2 - 3x)}$.

Factorizing,

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

Hence, $\bm{a^2b^2 + b^2 + a^2 + 1 = (a^2 + 1)(b^2 + 1)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow x^2 - 81 \\[1em] \Rightarrow (x)^2 - (9)^2 \\[1em] \Rightarrow (x + 9)(x - 9)\end{aligned}$$

Hence, $\bm{x^2 - 81 = (x + 9)(x - 9)}$.

Using the identity,

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

Factorizing,

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

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

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 36y^2 - 121 \\[1em] \Rightarrow (6y)^2 - (11)^2 \\[1em] \Rightarrow (6y + 11)(6y - 11)\end{aligned}$$

Hence, $\bm{36y^2 - 121 = (6y + 11)(6y - 11)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 49a^2 - 100b^2 \\[1em] \Rightarrow (7a)^2 - (10b)^2 \\[1em] \Rightarrow (7a + 10b)(7a - 10b)\end{aligned}$$

Hence, $\bm{49a^2 - 100b^2 = (7a + 10b)(7a - 10b)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow (a + b)^2 - 36 \\[1em] \Rightarrow (a + b)^2 - (6)^2 \\[1em] \Rightarrow [(a + b) + 6][(a + b) - 6] \\[1em] \Rightarrow (a + b + 6)(a + b - 6)\end{aligned}$$

Hence, $\bm{(a + b)^2 - 36 = (a + b + 6)(a + b - 6)}$.

Using the identity,

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

Factorizing,

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

Hence, $\bm{16c^2 - 1 = (4c + 1)(4c - 1)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 1 - 64b^2 \\[1em] \Rightarrow (1)^2 - (8b)^2 \\[1em] \Rightarrow (1 + 8b)(1 - 8b)\end{aligned}$$

Hence, $\bm{1 - 64b^2 = (1 + 8b)(1 - 8b)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow \frac{9}{16} - 25x^2 \\[1em] \Rightarrow \Big(\frac{3}{4}\Big)^2 - (5x)^2 \\[1em] \Rightarrow \Big(\frac{3}{4} + 5x\Big)\Big(\frac{3}{4} - 5x\Big)\end{aligned}$$

Hence, $\bm{\frac{9}{16} - 25x^2 = \Big(\frac{3}{4} + 5x\Big)\Big(\frac{3}{4} - 5x\Big)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow z^2 - \frac{1}{144} \\[1em] \Rightarrow (z)^2 - \Big(\frac{1}{12}\Big)^2 \\[1em] \Rightarrow \Big(z + \frac{1}{12}\Big)\Big(z - \frac{1}{12}\Big)\end{aligned}$$

Hence, $\bm{z^2 - \frac{1}{144} = \Big(z + \frac{1}{12}\Big)\Big(z - \frac{1}{12}\Big)}$.

Using the identity,

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

Factorizing,

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

Hence, $\bm{1 - (a - b)^2 = (1 + a - b)(1 - a + b)}$.

Using the identity,

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

Factorizing,

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

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

Using the identity,

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

Factorizing,

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

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

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 16(a + b)^2 - 9(a - b)^2 \\[1em] \Rightarrow [4(a + b)]^2 - [3(a - b)]^2 \\[1em] \Rightarrow [4(a + b) + 3(a - b)][4(a + b) - 3(a - b)] \\[1em] \Rightarrow (4a + 4b + 3a - 3b)(4a + 4b - 3a + 3b) \\[1em] \Rightarrow (7a + b)(a + 7b)\end{aligned}$$

Hence, $\bm{16(a + b)^2 - 9(a - b)^2 = (7a + b)(a + 7b)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 9(x + y)^2 - 16(x - 2y)^2 \\[1em] \Rightarrow [3(x + y)]^2 - [4(x - 2y)]^2 \\[1em] \Rightarrow [3(x + y) + 4(x - 2y)][3(x + y) - 4(x - 2y)] \\[1em] \Rightarrow (3x + 3y + 4x - 8y)(3x + 3y - 4x + 8y) \\[1em] \Rightarrow (7x - 5y)(11y - x)\end{aligned}$$

Hence, $\bm{9(x + y)^2 - 16(x - 2y)^2 = (7x - 5y)(11y - x)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 36(a - b)^2 - 25(a + b)^2 \\[1em] \Rightarrow [6(a - b)]^2 - [5(a + b)]^2 \\[1em] \Rightarrow [6(a - b) + 5(a + b)][6(a - b) - 5(a + b)] \\[1em] \Rightarrow (6a - 6b + 5a + 5b)(6a - 6b - 5a - 5b) \\[1em] \Rightarrow (11a - b)(a - 11b)\end{aligned}$$

Hence, $\bm{36(a - b)^2 - 25(a + b)^2 = (11a - b)(a - 11b)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 9(3x + 1)^2 - 4(x - 1)^2 \\[1em] \Rightarrow [3(3x + 1)]^2 - [2(x - 1)]^2 \\[1em] \Rightarrow [3(3x + 1) + 2(x - 1)][3(3x + 1) - 2(x - 1)] \\[1em] \Rightarrow (9x + 3 + 2x - 2)(9x + 3 - 2x + 2) \\[1em] \Rightarrow (11x + 1)(7x + 5)\end{aligned}$$

Hence, $\bm{9(3x + 1)^2 - 4(x - 1)^2 = (11x + 1)(7x + 5)}$.

Using the identity,

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

Factorizing,

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

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

Using the identity,

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

Factorizing,

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

Hence, $\bm{x^2 - a^2 - 2a - 1 = (x + a + 1)(x - a - 1)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow x^2 - m^2 + 6mn - 9n^2 \\[1em] \Rightarrow x^2 - (m^2 - 6mn + 9n^2) \\[1em] \Rightarrow (x)^2 - (m - 3n)^2 \\[1em] \Rightarrow [x + (m - 3n)][x - (m - 3n)] \\[1em] \Rightarrow (x + m - 3n)(x - m + 3n)\end{aligned}$$

Hence, $\bm{x^2 - m^2 + 6mn - 9n^2 = (x + m - 3n)(x - m + 3n)}$.

Using the identity,

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

Factorizing,

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

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

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 16a^4 - 81b^4 \\[1em] \Rightarrow (4a^2)^2 - (9b^2)^2 \\[1em] \Rightarrow (4a^2 + 9b^2)(4a^2 - 9b^2) \\[1em] \Rightarrow (4a^2 + 9b^2)[(2a)^2 - (3b)^2] \\[1em] \Rightarrow (4a^2 + 9b^2)(2a + 3b)(2a - 3b)\end{aligned}$$

Hence, $\bm{16a^4 - 81b^4 = (4a^2 + 9b^2)(2a + 3b)(2a - 3b)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 3 - 75z^2 \\[1em] \Rightarrow 3(1 - 25z^2) \\[1em] \Rightarrow 3[(1)^2 - (5z)^2] \\[1em] \Rightarrow 3(1 + 5z)(1 - 5z)\end{aligned}$$

Hence, $\bm{3 - 75z^2 = 3(1 + 5z)(1 - 5z)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 48a^2b^2 - 3 \\[1em] \Rightarrow 3(16a^2b^2 - 1) \\[1em] \Rightarrow 3[(4ab)^2 - (1)^2] \\[1em] \Rightarrow 3(4ab + 1)(4ab - 1)\end{aligned}$$

Hence, $\bm{48a^2b^2 - 3 = 3(4ab + 1)(4ab - 1)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 4x^3 - 81x \\[1em] \Rightarrow x(4x^2 - 81) \\[1em] \Rightarrow x[(2x)^2 - (9)^2] \\[1em] \Rightarrow x(2x + 9)(2x - 9)\end{aligned}$$

Hence, $\bm{4x^3 - 81x = x(2x + 9)(2x - 9)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 9b^3 - 144b \\[1em] \Rightarrow 9b(b^2 - 16) \\[1em] \Rightarrow 9b[(b)^2 - (4)^2] \\[1em] \Rightarrow 9b(b + 4)(b - 4)\end{aligned}$$

Hence, $\bm{9b^3 - 144b = 9b(b + 4)(b - 4)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 32x^2 - 72y^2 \\[1em] \Rightarrow 8(4x^2 - 9y^2) \\[1em] \Rightarrow 8[(2x)^2 - (3y)^2] \\[1em] \Rightarrow 8(2x + 3y)(2x - 3y)\end{aligned}$$

Hence, $\bm{32x^2 - 72y^2 = 8(2x + 3y)(2x - 3y)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 50x^2y - 32y^3 \\[1em] \Rightarrow 2y(25x^2 - 16y^2) \\[1em] \Rightarrow 2y[(5x)^2 - (4y)^2] \\[1em] \Rightarrow 2y(5x + 4y)(5x - 4y)\end{aligned}$$

Hence, $\bm{50x^2y - 32y^3 = 2y(5x + 4y)(5x - 4y)}$.

Using the identity,

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

Factorizing,

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

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

Using the identity,

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

Factorizing,

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

Hence, $\bm{ab^3c - abc^3 = abc(b + c)(b - c)}$.

Using the identity,

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

Factorizing,

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

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

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 1 - 0.49c^6 \\[1em] \Rightarrow (1)^2 - (0.7c^3)^2 \\[1em] \Rightarrow (1 + 0.7c^3)(1 - 0.7c^3)\end{aligned}$$

Hence, $\bm{1 - 0.49c^6 = (1 + 0.7c^3)(1 - 0.7c^3)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow x^2 - y^2 - 8yz - 16z^2 \\[1em] \Rightarrow x^2 - (y^2 + 8yz + 16z^2) \\[1em] \Rightarrow (x)^2 - (y + 4z)^2 \\[1em] \Rightarrow [x + (y + 4z)][x - (y + 4z)] \\[1em] \Rightarrow (x + y + 4z)(x - y - 4z)\end{aligned}$$

Hence, $\bm{x^2 - y^2 - 8yz - 16z^2 = (x + y + 4z)(x - y - 4z)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow x^3y^3 - \frac{25xy}{z^2} \\[1em] \Rightarrow xy\Big(x^2y^2 - \frac{25}{z^2}\Big) \\[1em] \Rightarrow xy\Big[(xy)^2 - \Big(\frac{5}{z}\Big)^2\Big] \\[1em] \Rightarrow xy\Big(xy + \frac{5}{z}\Big)\Big(xy - \frac{5}{z}\Big)\end{aligned}$$

Hence, $\bm{x^3y^3 - \frac{25xy}{z^2} = xy\Big(xy + \frac{5}{z}\Big)\Big(xy - \frac{5}{z}\Big)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 0.0324x^4 - 0.0064b^4 \\[1em] \Rightarrow 4(0.0081x^4 - 0.0016b^4) \\[1em] \Rightarrow 4[(0.09x^2)^2 - (0.04b^2)^2] \\[1em] \Rightarrow 4(0.09x^2 + 0.04b^2)(0.09x^2 - 0.04b^2) \\[1em] \Rightarrow 4(0.09x^2 + 0.04b^2)[(0.3x)^2 - (0.2b)^2] \\[1em] \Rightarrow 4(0.09x^2 + 0.04b^2)(0.3x + 0.2b)(0.3x - 0.2b)\end{aligned}$$

Hence, $\bm{0.0324x^4 - 0.0064b^4 = 4(0.09x^2 + 0.04b^2)(0.3x + 0.2b)(0.3x - 0.2b)}$.

Using the identity,

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

⇒ 82² - 18²

⇒ (82 + 18)(82 - 18)

⇒ 100 × 64

⇒ 6400

Hence, 82² - 18² = 6400.

Using the identity,

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

⇒ (15.8)² - (9.2)²

⇒ (15.8 + 9.2)(15.8 - 9.2)

⇒ 25 × 6.6

⇒ 165

Hence, (15.8)² - (9.2)² = 165.

Using the identity,

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

⇒ (0.8)² - (0.2)²

⇒ (0.8 + 0.2)(0.8 - 0.2)

⇒ 1 × 0.6

⇒ 0.6

Hence, (0.8)² - (0.2)² = 0.6.

Using the identity,

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

$$\begin{aligned}\Rightarrow \left(7\frac{3}{4}\right)^2 - \left(2\frac{1}{4}\right)^2 \\[1em] \Rightarrow \Big(7\frac{3}{4} + 2\frac{1}{4}\Big)\Big(7\frac{3}{4} - 2\frac{1}{4}\Big) \\[1em] \Rightarrow \Big(\frac{31}{4} + \frac{9}{4}\Big)\Big(\frac{31}{4} - \frac{9}{4}\Big) \\[1em] \Rightarrow \frac{40}{4} \times \frac{22}{4} \\[1em] \Rightarrow 10 \times \frac{11}{2} \\[1em] \Rightarrow 55\end{aligned}$$

Hence, $\mathbf{\left(7\frac{3}{4}\right)^2 - \left(2\frac{1}{4}\right)^2 = 55}$.

Using the identity,

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

$$\begin{aligned}\Rightarrow \left(6\frac{4}{11}\right)^2 - \left(4\frac{7}{11}\right)^2 \\[1em] \Rightarrow \Big(6\frac{4}{11} + 4\frac{7}{11}\Big)\Big(6\frac{4}{11} - 4\frac{7}{11}\Big) \\[1em] \Rightarrow \Big(\frac{70}{11} + \frac{51}{11}\Big)\Big(\frac{70}{11} - \frac{51}{11}\Big) \\[1em] \Rightarrow \frac{121}{11} \times \frac{19}{11} \\[1em] \Rightarrow 11 \times \frac{19}{11} \\[1em] \Rightarrow 19\end{aligned}$$

Hence, $\mathbf{\left(6\frac{4}{11}\right)^2 - \left(4\frac{7}{11}\right)^2 = 19}$.

Using the identity,

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

$$\begin{aligned}\Rightarrow \frac{(7.3)^2 - (2.7)^2}{7.3 - 2.7} \\[1em] \Rightarrow \frac{(7.3 + 2.7)(7.3 - 2.7)}{7.3 - 2.7} \\[1em] \Rightarrow 7.3 + 2.7 \\[1em] \Rightarrow 10\end{aligned}$$

Hence, $\mathbf{\frac{7.3 \times 7.3 - 2.7 \times 2.7}{7.3 - 2.7} = 10}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow x^2 + 18x + 81 \\[1em] \Rightarrow (x)^2 + 2 \times x \times 9 + (9)^2 \\[1em] \Rightarrow (x + 9)^2\end{aligned}$$

Hence, $\bm{x^2 + 18x + 81 = (x + 9)^2}$.

Using the identity,

[$(x - y)^2 = x^2 - 2xy + y^2$]

Factorizing,

$$\begin{aligned}\Rightarrow a^2 - 14a + 49 \\[1em] \Rightarrow (a)^2 - 2 \times a \times 7 + (7)^2 \\[1em] \Rightarrow (a - 7)^2\end{aligned}$$

Hence, $\bm{a^2 - 14a + 49 = (a - 7)^2}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 4x^2 + 12x + 9 \\[1em] \Rightarrow (2x)^2 + 2 \times 2x \times 3 + (3)^2 \\[1em] \Rightarrow (2x + 3)^2\end{aligned}$$

Hence, $\bm{4x^2 + 12x + 9 = (2x + 3)^2}$.

Using the identity,

[$(x - y)^2 = x^2 - 2xy + y^2$]

Factorizing,

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

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

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 16 + 56x + 49x^2 \\[1em] \Rightarrow (4)^2 + 2 \times 4 \times 7x + (7x)^2 \\[1em] \Rightarrow (4 + 7x)^2\end{aligned}$$

Hence, $\bm{16 + 56x + 49x^2 = (4 + 7x)^2}$.

Using the identity,

[$(x - y)^2 = x^2 - 2xy + y^2$]

Factorizing,

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

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

Using the identity,

[$(x - y)^2 = x^2 - 2xy + y^2$]

Factorizing,

$$\begin{aligned}\Rightarrow 9q^4r^4 - 6p^4q^2r^2 + p^8 \\[1em] \Rightarrow (3q^2r^2)^2 - 2 \times 3q^2r^2 \times p^4 + (p^4)^2 \\[1em] \Rightarrow (3q^2r^2 - p^4)^2\end{aligned}$$

Hence, $\bm{9q^4r^4 - 6p^4q^2r^2 + p^8 = (3q^2r^2 - p^4)^2}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow \frac{9p^2}{q^2} + \frac{16r^2}{m^2} + \frac{24pr}{qm} \\[1em] \Rightarrow \Big(\frac{3p}{q}\Big)^2 + 2 \times \frac{3p}{q} \times \frac{4r}{m} + \Big(\frac{4r}{m}\Big)^2 \\[1em] \Rightarrow \Big(\frac{3p}{q} + \frac{4r}{m}\Big)^2\end{aligned}$$

Hence, $\bm{\frac{9p^2}{q^2} + \frac{16r^2}{m^2} + \frac{24pr}{qm} = \Big(\frac{3p}{q} + \frac{4r}{m}\Big)^2}$.

Using the identity,

[$(x - y)^2 = x^2 - 2xy + y^2$]

Factorizing,

$$\begin{aligned}\Rightarrow \frac{1}{4}z^6 + 9a^2 - 3az^3 \\[1em] \Rightarrow \Big(\frac{1}{2}z^3\Big)^2 - 2 \times \frac{1}{2}z^3 \times 3a + (3a)^2 \\[1em] \Rightarrow \Big(\frac{1}{2}z^3 - 3a\Big)^2\end{aligned}$$

Hence, $\bm{\frac{1}{4}z^6 + 9a^2 - 3az^3 = \Big(\frac{1}{2}z^3 - 3a\Big)^2}$.

Using the identity,

[$(x - y)^2 = x^2 - 2xy + y^2$]

Factorizing,

$$\begin{aligned}\Rightarrow \frac{9}{4}a^2 + \frac{49}{9}p^2 - 7ap \\[1em] \Rightarrow \Big(\frac{3}{2}a\Big)^2 - 2 \times \frac{3}{2}a \times \frac{7}{3}p + \Big(\frac{7}{3}p\Big)^2 \\[1em] \Rightarrow \Big(\frac{3}{2}a - \frac{7}{3}p\Big)^2\end{aligned}$$

Hence, $\bm{\frac{9}{4}a^2 + \frac{49}{9}p^2 - 7ap = \Big(\frac{3}{2}a - \frac{7}{3}p\Big)^2}$.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is 22 and product is 85.

By trial, we see that 5 + 17 = 22 and 5 × 17 = 85.

Factorizing,

$$\begin{aligned}\Rightarrow t^2 + 22t + 85 \\[1em] \Rightarrow t^2 + (5 + 17)t + 85 \\[1em] \Rightarrow t^2 + 5t + 17t + 85 \\[1em] \Rightarrow t(t + 5) + 17(t + 5) \\[1em] \Rightarrow (t + 5)(t + 17)\end{aligned}$$

Hence, $\bm{t^2 + 22t + 85 = (t + 5)(t + 17)}$.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is -10 and product is 24.

By trial, we see that (-6) + (-4) = -10 and (-6) × (-4) = 24.

Factorizing,

$$\begin{aligned}\Rightarrow x^2 - 10x + 24 \\[1em] \Rightarrow x^2 - (6 + 4)x + 24 \\[1em] \Rightarrow x^2 - 6x - 4x + 24 \\[1em] \Rightarrow x(x - 6) - 4(x - 6) \\[1em] \Rightarrow (x - 6)(x - 4)\end{aligned}$$

Hence, $\bm{x^2 - 10x + 24 = (x - 6)(x - 4)}$.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is -3 and product is -40.

By trial, we see that (-8) + 5 = -3 and (-8) × 5 = -40.

Factorizing,

$$\begin{aligned}\Rightarrow m^2 - 3m - 40 \\[1em] \Rightarrow m^2 - (8 - 5)m - 40 \\[1em] \Rightarrow m^2 - 8m + 5m - 40 \\[1em] \Rightarrow m(m - 8) + 5(m - 8) \\[1em] \Rightarrow (m - 8)(m + 5)\end{aligned}$$

Hence, $\bm{m^2 - 3m - 40 = (m - 8)(m + 5)}$.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is 1 and product is -72.

By trial, we see that 9 + (-8) = 1 and 9 × (-8) = -72.

Factorizing,

$$\begin{aligned}\Rightarrow x^2 + x - 72 \\[1em] \Rightarrow x^2 + (9 - 8)x - 72 \\[1em] \Rightarrow x^2 + 9x - 8x - 72 \\[1em] \Rightarrow x(x + 9) - 8(x + 9) \\[1em] \Rightarrow (x + 9)(x - 8)\end{aligned}$$

Hence, $\bm{x^2 + x - 72 = (x + 9)(x - 8)}$.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is -7 and product is -120.

By trial, we see that (-15) + 8 = -7 and (-15) × 8 = -120.

Factorizing,

$$\begin{aligned}\Rightarrow p^2 - 7p - 120 \\[1em] \Rightarrow p^2 - (15 - 8)p - 120 \\[1em] \Rightarrow p^2 - 15p + 8p - 120 \\[1em] \Rightarrow p(p - 15) + 8(p - 15) \\[1em] \Rightarrow (p - 15)(p + 8)\end{aligned}$$

Hence, $\bm{p^2 - 7p - 120 = (p - 15)(p + 8)}$.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is -17 and product is 16.

By trial, we see that (-1) + (-16) = -17 and (-1) × (-16) = 16.

Factorizing,

$$\begin{aligned}\Rightarrow 16 - 17z + z^2 \\[1em] \Rightarrow 16 - (1 + 16)z + z^2 \\[1em] \Rightarrow 16 - z - 16z + z^2 \\[1em] \Rightarrow (16 - z) - (16z - z^2) \\[1em] \Rightarrow 1(16 - z) - z(16 - z) \\[1em] \Rightarrow (16 - z)(1 - z)\end{aligned}$$

Hence, $\bm{16 - 17z + z^2 = (16 - z)(1 - z)}$.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is 5 and product is -104.

By trial, we see that 13 + (-8) = 5 and 13 × (-8) = -104.

Factorizing,

$$\begin{aligned}\Rightarrow a^2 + 5a - 104 \\[1em] \Rightarrow a^2 + (13 - 8)a - 104 \\[1em] \Rightarrow a^2 + 13a - 8a - 104 \\[1em] \Rightarrow a(a + 13) - 8(a + 13) \\[1em] \Rightarrow (a + 13)(a - 8)\end{aligned}$$

Hence, $\bm{a^2 + 5a - 104 = (a + 13)(a - 8)}$.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is 11 and product is 3 × 10 = 30.

By trial, we see that 6 + 5 = 11 and 6 × 5 = 30.

Factorizing,

$$\begin{aligned}\Rightarrow 3x^2 + 11x + 10 \\[1em] \Rightarrow 3x^2 + (6 + 5)x + 10 \\[1em] \Rightarrow 3x^2 + 6x + 5x + 10 \\[1em] \Rightarrow 3x(x + 2) + 5(x + 2) \\[1em] \Rightarrow (x + 2)(3x + 5)\end{aligned}$$

Hence, $\bm{3x^2 + 11x + 10 = (3x + 5)(x + 2)}$.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is 7 and product is 6 × (-3) = -18.

By trial, we see that 9 + (-2) = 7 and 9 × (-2) = -18.

Factorizing,

$$\begin{aligned}\Rightarrow 6x^2 + 7x - 3 \\[1em] \Rightarrow 6x^2 + (9 - 2)x - 3 \\[1em] \Rightarrow 6x^2 + 9x - 2x - 3 \\[1em] \Rightarrow 3x(2x + 3) - 1(2x + 3) \\[1em] \Rightarrow (2x + 3)(3x - 1)\end{aligned}$$

Hence, $\bm{6x^2 + 7x - 3 = (2x + 3)(3x - 1)}$.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is -4 and product is 3 × (-4) = -12.

By trial, we see that (-6) + 2 = -4 and (-6) × 2 = -12.

Factorizing,

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

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

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is -1 and product is 72 × (-1) = -72.

By trial, we see that 8 + (-9) = -1 and 8 × (-9) = -72.

Factorizing,

$$\begin{aligned}\Rightarrow 72 - x - x^2 \\[1em] \Rightarrow 72 + (8 - 9)x - x^2 \\[1em] \Rightarrow 72 + 8x - 9x - x^2 \\[1em] \Rightarrow 8(9 + x) - x(9 + x) \\[1em] \Rightarrow (9 + x)(8 - x)\end{aligned}$$

Hence, $\bm{72 - x - x^2 = (8 - x)(9 + x)}$.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is -3 and product is -40.

By trial, we see that (-8) + 5 = -3 and (-8) × 5 = -40.

Factorizing,

$$\begin{aligned}\Rightarrow x^2 - 3xy - 40y^2 \\[1em] \Rightarrow x^2 - (8 - 5)xy - 40y^2 \\[1em] \Rightarrow x^2 - 8xy + 5xy - 40y^2 \\[1em] \Rightarrow x(x - 8y) + 5y(x - 8y) \\[1em] \Rightarrow (x - 8y)(x + 5y)\end{aligned}$$

Hence, $\bm{x^2 - 3xy - 40y^2 = (x + 5y)(x - 8y)}$.

Here $y$ is a common factor, so we first take $y$ outside.

$$\begin{aligned}\Rightarrow 3x^2y + 11xy + 6y \\[1em] \Rightarrow y(3x^2 + 11x + 6)\end{aligned}$$

The expression left inside the bracket is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is 11 and product is 3 × 6 = 18.

By trial, we see that 9 + 2 = 11 and 9 × 2 = 18.

Factorizing,

$$\begin{aligned}\Rightarrow y(3x^2 + 11x + 6) \\[1em] \Rightarrow y(3x^2 + 9x + 2x + 6) \\[1em] \Rightarrow y[3x(x + 3) + 2(x + 3)] \\[1em] \Rightarrow y(x + 3)(3x + 2)\end{aligned}$$

Hence, $\bm{3x^2y + 11xy + 6y = y(x + 3)(3x + 2)}$.

We first substitute a single variable for the repeated expression, so that the given expression becomes a trinomial. Then we factorize it by splitting the middle term.

Let $(a - b) = x$, then the given expression becomes $x^2 - 5x + 6$.

We want to find two integers whose sum is -5 and product is 6.

By trial, we see that (-2) + (-3) = -5 and (-2) × (-3) = 6.

Factorizing,

$$\begin{aligned}\Rightarrow x^2 - 5x + 6 \\[1em] \Rightarrow x^2 - 2x - 3x + 6 \\[1em] \Rightarrow x(x - 2) - 3(x - 2) \\[1em] \Rightarrow (x - 2)(x - 3)\end{aligned}$$

Substituting $x = (a - b)$, we get

⇒ $(a - b - 2)(a - b - 3)$

Hence, $\bm{(a - b)^2 - 5(a - b) + 6 = (a - b - 2)(a - b - 3)}$.

We first substitute a single variable for the repeated expression, so that the given expression becomes a trinomial. Then we factorize it by splitting the middle term.

Let $(a - 3b) = x$, then the given expression becomes $x^2 - 4x - 21$.

We want to find two integers whose sum is -4 and product is -21.

By trial, we see that (-7) + 3 = -4 and (-7) × 3 = -21.

Factorizing,

$$\begin{aligned}\Rightarrow x^2 - 4x - 21 \\[1em] \Rightarrow x^2 - 7x + 3x - 21 \\[1em] \Rightarrow x(x - 7) + 3(x - 7) \\[1em] \Rightarrow (x - 7)(x + 3)\end{aligned}$$

Substituting $x = (a - 3b)$, we get

⇒ $(a - 3b - 7)(a - 3b + 3)$

Hence, $\bm{(a - 3b)^2 - 4(a - 3b) - 21 = (a - 3b - 7)(a - 3b + 3)}$.

We first substitute a single variable for the repeated expression, so that the given expression becomes a trinomial. Then we factorize it by splitting the middle term.

Let $(y - 2) = x$, then the given expression becomes $3x^2 - x - 44$.

We want to find two integers whose sum is -1 and product is 3 × (-44) = -132.

By trial, we see that (-12) + 11 = -1 and (-12) × 11 = -132.

Factorizing,

$$\begin{aligned}\Rightarrow 3x^2 - x - 44 \\[1em] \Rightarrow 3x^2 - 12x + 11x - 44 \\[1em] \Rightarrow 3x(x - 4) + 11(x - 4) \\[1em] \Rightarrow (x - 4)(3x + 11)\end{aligned}$$

Substituting $x = (y - 2)$, we get

$$\begin{aligned}\Rightarrow (y - 2 - 4)[3(y - 2) + 11] \\[1em] \Rightarrow (y - 6)(3y - 6 + 11) \\[1em] \Rightarrow (y - 6)(3y + 5)\end{aligned}$$

Hence, $\bm{3(y - 2)^2 - (y - 2) - 44 = (y - 6)(3y + 5)}$.

We first substitute a single variable for the repeated expression, so that the given expression becomes a trinomial. Then we factorize it by splitting the middle term.

Let $(x + y) = a$, then the given expression becomes $7 + 10a - 8a^2$.

We want to find two integers whose sum is 10 and product is 7 × (-8) = -56.

By trial, we see that 14 + (-4) = 10 and 14 × (-4) = -56.

Factorizing,

$$\begin{aligned}\Rightarrow 7 + 10a - 8a^2 \\[1em] \Rightarrow 7 + 14a - 4a - 8a^2 \\[1em] \Rightarrow 7(1 + 2a) - 4a(1 + 2a) \\[1em] \Rightarrow (1 + 2a)(7 - 4a)\end{aligned}$$

Substituting $a = (x + y)$, we get

$$\begin{aligned}\Rightarrow [1 + 2(x + y)][7 - 4(x + y)] \\[1em] \Rightarrow (1 + 2x + 2y)(7 - 4x - 4y)\end{aligned}$$

Hence, $\bm{7 + 10(x + y) - 8(x + y)^2 = (1 + 2x + 2y)(7 - 4x - 4y)}$.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 7x^2 - 63y^2 \\[1em] \Rightarrow 7(x^2 - 9y^2) \\[1em] \Rightarrow 7[(x)^2 - (3y)^2] \\[1em] \Rightarrow 7(x + 3y)(x - 3y)\end{aligned}$$

Hence, option 4 is the correct option.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 2p - 32p^3 \\[1em] \Rightarrow 2p(1 - 16p^2) \\[1em] \Rightarrow 2p[(1)^2 - (4p)^2] \\[1em] \Rightarrow 2p(1 + 4p)(1 - 4p)\end{aligned}$$

Hence, option 3 is the correct option.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow a^3 - 144a \\[1em] \Rightarrow a(a^2 - 144) \\[1em] \Rightarrow a[(a)^2 - (12)^2] \\[1em] \Rightarrow a(a + 12)(a - 12)\end{aligned}$$

Hence, option 2 is the correct option.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 2 - 50z^2 \\[1em] \Rightarrow 2(1 - 25z^2) \\[1em] \Rightarrow 2[(1)^2 - (5z)^2] \\[1em] \Rightarrow 2(1 + 5z)(1 - 5z)\end{aligned}$$

Hence, option 3 is the correct option.

Factorizing,

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

Hence, option 4 is the correct option.

Factorizing,

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

Hence, option 1 is the correct option.

Factorizing,

$$\begin{aligned}\Rightarrow xy - pq + xq - yp \\[1em] \Rightarrow (xy + xq) - (yp + pq) \\[1em] \Rightarrow x(y + q) - p(y + q) \\[1em] \Rightarrow (y + q)(x - p)\end{aligned}$$

Hence, option 3 is the correct option.

Factorizing,

$$\begin{aligned}\Rightarrow mn - m - n + 1 \\[1em] \Rightarrow (mn - m) - (n - 1) \\[1em] \Rightarrow m(n - 1) - 1(n - 1) \\[1em] \Rightarrow (m - 1)(n - 1)\end{aligned}$$

Hence, option 4 is the correct option.

Factorizing,

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

Hence, option 2 is the correct option.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow 12a^2 - 27 \\[1em] \Rightarrow 3(4a^2 - 9) \\[1em] \Rightarrow 3[(2a)^2 - (3)^2] \\[1em] \Rightarrow 3(2a + 3)(2a - 3)\end{aligned}$$

Hence, option 1 is the correct option.

Using the identity,

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

Factorizing,

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

Hence, option 4 is the correct option.

Using the identity,

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

Factorizing,

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

Hence, option 2 is the correct option.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is 6 and product is 8.

By trial, we see that 2 + 4 = 6 and 2 × 4 = 8.

Factorizing,

$$\begin{aligned}\Rightarrow a^2 + 6a + 8 \\[1em] \Rightarrow a^2 + (2 + 4)a + 8 \\[1em] \Rightarrow a^2 + 2a + 4a + 8 \\[1em] \Rightarrow a(a + 2) + 4(a + 2) \\[1em] \Rightarrow (a + 2)(a + 4)\end{aligned}$$

Hence, option 2 is the correct option.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is 4 and product is -21.

By trial, we see that 7 + (-3) = 4 and 7 × (-3) = -21.

Factorizing,

$$\begin{aligned}\Rightarrow p^2 + 4p - 21 \\[1em] \Rightarrow p^2 + (7 - 3)p - 21 \\[1em] \Rightarrow p^2 + 7p - 3p - 21 \\[1em] \Rightarrow p(p + 7) - 3(p + 7) \\[1em] \Rightarrow (p + 7)(p - 3)\end{aligned}$$

Hence, option 3 is the correct option.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is 2 and product is -3.

By trial, we see that 3 + (-1) = 2 and 3 × (-1) = -3.

Factorizing,

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

Hence, option 4 is the correct option.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is 3 and product is 40 × (-1) = -40.

By trial, we see that 8 + (-5) = 3 and 8 × (-5) = -40.

Factorizing,

$$\begin{aligned}\Rightarrow 40 + 3y - y^2 \\[1em] \Rightarrow 40 + (8 - 5)y - y^2 \\[1em] \Rightarrow 40 + 8y - 5y - y^2 \\[1em] \Rightarrow 8(5 + y) - y(5 + y) \\[1em] \Rightarrow (5 + y)(8 - y)\end{aligned}$$

Hence, option 2 is the correct option.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is 5 and product is 2 × 3 = 6.

By trial, we see that 2 + 3 = 5 and 2 × 3 = 6.

Factorizing,

$$\begin{aligned}\Rightarrow 2a^2 + 5a + 3 \\[1em] \Rightarrow 2a^2 + (2 + 3)a + 3 \\[1em] \Rightarrow 2a^2 + 2a + 3a + 3 \\[1em] \Rightarrow 2a(a + 1) + 3(a + 1) \\[1em] \Rightarrow (a + 1)(2a + 3)\end{aligned}$$

Hence, option 3 is the correct option.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is -13 and product is 6 × 6 = 36.

By trial, we see that (-4) + (-9) = -13 and (-4) × (-9) = 36.

Factorizing,

$$\begin{aligned}\Rightarrow 6x^2 - 13x + 6 \\[1em] \Rightarrow 6x^2 - (4 + 9)x + 6 \\[1em] \Rightarrow 6x^2 - 4x - 9x + 6 \\[1em] \Rightarrow 2x(3x - 2) - 3(3x - 2) \\[1em] \Rightarrow (3x - 2)(2x - 3)\end{aligned}$$

Hence, option 4 is the correct option.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is -8 and product is 4 × 3 = 12.

By trial, we see that (-2) + (-6) = -8 and (-2) × (-6) = 12.

Factorizing,

$$\begin{aligned}\Rightarrow 4p^2 - 8p + 3 \\[1em] \Rightarrow 4p^2 - (2 + 6)p + 3 \\[1em] \Rightarrow 4p^2 - 2p - 6p + 3 \\[1em] \Rightarrow 2p(2p - 1) - 3(2p - 1) \\[1em] \Rightarrow (2p - 1)(2p - 3)\end{aligned}$$

Hence, option 4 is the correct option.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is 23 and product is 3 × (-8) = -24.

By trial, we see that 24 + (-1) = 23 and 24 × (-1) = -24.

Factorizing,

$$\begin{aligned}\Rightarrow 3 + 23x - 8x^2 \\[1em] \Rightarrow 3 + (24 - 1)x - 8x^2 \\[1em] \Rightarrow 3 + 24x - x - 8x^2 \\[1em] \Rightarrow 3(1 + 8x) - x(1 + 8x) \\[1em] \Rightarrow (1 + 8x)(3 - x)\end{aligned}$$

Hence, option 3 is the correct option.

Factorising each of the four given expressions, we get

$$\begin{aligned}\Rightarrow A = x^2 + x - 12 = x^2 + 4x - 3x - 12 = x(x + 4) - 3(x + 4) = (x + 4)(x - 3) \\[1em] \Rightarrow B = x^2 - 9 = (x)^2 - (3)^2 = (x + 3)(x - 3) \\[1em] \Rightarrow C = x^2 - 8x + 16 = (x)^2 - 2 \times x \times 4 + (4)^2 = (x - 4)^2 \\[1em] \Rightarrow D = x^2 + 6x + 9 = (x)^2 + 2 \times x \times 3 + (3)^2 = (x + 3)^2\end{aligned}$$

A and B have the common factor $(x - 3)$.

B and D have the common factor $(x + 3)$.

A has the factors $(x + 4)$ and $(x - 3)$, while C has the factor $(x - 4)$ only. So, A and C have no common factor.

Hence, option 2 is the correct option.

From the factorisation done above, we have

$$\begin{aligned}\Rightarrow A = (x + 4)(x - 3) \\[1em] \Rightarrow B = (x + 3)(x - 3) \\[1em] \Rightarrow C = (x - 4)^2 \\[1em] \Rightarrow D = (x + 3)^2\end{aligned}$$

$(x - 3)$ occurs as a factor of A and of B, that is, of two expressions.

Hence, option 3 is the correct option.

Expression A $= x^2 + x - 12 = (x + 4)(x - 3)$

Factorising each of the given options, we get

$$\begin{aligned}\Rightarrow x^2 - 2x - 15 = x^2 - 5x + 3x - 15 = (x - 5)(x + 3) \\[1em] \Rightarrow x^2 - 9x + 20 = x^2 - 5x - 4x + 20 = (x - 5)(x - 4) \\[1em] \Rightarrow x^2 + x - 20 = x^2 + 5x - 4x - 20 = (x + 5)(x - 4) \\[1em] \Rightarrow x^2 - 16 = (x)^2 - (4)^2 = (x + 4)(x - 4)\end{aligned}$$

Only $x^2 - 16$ has the factor $(x + 4)$, which is also a factor of A.

Hence, option 4 is the correct option.

Expression C $= x^2 - 8x + 16 = (x - 4)^2 = (x - 4)(x - 4)$

Factorising each of the given options, we get

⇒ $x^2 + 2x + 8$ cannot be factorised, as there are no two integers whose sum is 2 and product is 8.

$$\begin{aligned}\Rightarrow x^2 - 7x + 12 = x^2 - 3x - 4x + 12 = (x - 3)(x - 4) \\[1em] \Rightarrow x^2 - x - 20 = x^2 - 5x + 4x - 20 = (x - 5)(x + 4) \\[1em] \Rightarrow x^2 + 3x - 4 = x^2 + 4x - x - 4 = (x + 4)(x - 1)\end{aligned}$$

Only $x^2 - 7x + 12$ has the factor $(x - 4)$, which is also a factor of C.

Hence, option 2 is the correct option.

We first factorize $x^2 - 2xy + y^2$ to check whether Assertion (A) is true.

Using the identity,

[$(x - y)^2 = x^2 - 2xy + y^2$]

Factorizing,

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

So, the factors of $x^2 - 2xy + y^2$ are $(x - y)$ and $(x - y)$, and not $(x + y)$ and $(x - y)$.

So, Assertion (A) is false.

$(x + y)^2 = x^2 + 2xy + y^2$ is a standard algebraic identity.

So, Reason (R) is true.

Hence, option 4 is the correct option.

We first factorize $lx + my + mx + ly$ by grouping the terms to check whether Assertion (A) is true.

Factorizing,

$$\begin{aligned}\Rightarrow lx + my + mx + ly \\[1em] \Rightarrow (lx + ly) + (mx + my) \\[1em] \Rightarrow l(x + y) + m(x + y) \\[1em] \Rightarrow (x + y)(l + m)\end{aligned}$$

So, Assertion (A) is true.

When an algebraic expression can be written as the product of two or more expressions, then each of these expressions is called a factor of the given expression. This is the definition of a factor, and it is exactly what tells us that $(l + m)$ and $(x + y)$ are the factors of $lx + my + mx + ly$.

So, Reason (R) is true and it is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

Factorizing,

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

Hence, option 1 is the correct option.

Using the identity,

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

Factorizing,

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

Hence, option 4 is the correct option.

An expression is said to be in factorised form when it is written as the product of two or more expressions.

$(2ab + 5)(a + 2)$ is the product of $(2ab + 5)$ and $(a + 2)$, so it is in factorised form.

In $5a + 2ab + 14$, the terms $5a$, $2ab$ and 14 have no common factor, so the expression cannot be factorised. It is already in its simplest form.

$5(a + 3)$ is the product of 5 and $(a + 3)$, so it is in factorised form.

$2a^2 + 9(a - 2)$ is the sum of the two terms $2a^2$ and $9(a - 2)$, and it can be factorised further as shown below.

$$\begin{aligned}\Rightarrow 2a^2 + 9(a - 2) \\[1em] \Rightarrow 2a^2 + 9a - 18 \\[1em] \Rightarrow 2a^2 + (12 - 3)a - 18 \\[1em] \Rightarrow 2a^2 + 12a - 3a - 18 \\[1em] \Rightarrow 2a(a + 6) - 3(a + 6) \\[1em] \Rightarrow (a + 6)(2a - 3)\end{aligned}$$

So, $2a^2 + 9(a - 2)$ is not in factorised form.

Hence, option 4 is the correct option.

In Step I, the three terms have been written as

$$\begin{aligned}\Rightarrow 5 \times 2 \times 2 \times x \times x = 20x^2 \\[1em] \Rightarrow 5 \times x \times y \times y = 5xy^2 \\[1em] \Rightarrow 5 \times 5 \times x \times y \times z = 25xyz\end{aligned}$$

So, Step I is correct.

In Step II, the common factor $5x$ is taken out of each term. As the third term of the given expression is $-25xyz$, the third term inside the bracket must be $-5 \times y \times z$ and not $+5 \times y \times z$.

The correct Step II is $5 \times x (2 \times 2 + y \times y - 5 \times y \times z)$.

So, the student made his first mistake in Step II.

Hence, option 2 is the correct option.

Taking the regrouping given in option 1, we get

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

Taking the regrouping given in option 2, we get

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

Both the regroupings factorise the given expression and give the same factors.

Hence, option 3 is the correct option.

The given expression is a trinomial, so we factorize it by splitting the middle term.

We want to find two integers whose sum is -11 and product is 3 × 6 = 18.

By trial, we see that (-9) + (-2) = -11 and (-9) × (-2) = 18.

Factorizing,

$$\begin{aligned}\Rightarrow 3x^2 - 11x + 6 \\[1em] \Rightarrow 3x^2 - (9 + 2)x + 6 \\[1em] \Rightarrow 3x^2 - 9x - 2x + 6 \\[1em] \Rightarrow 3x(x - 3) - 2(x - 3) \\[1em] \Rightarrow (x - 3)(3x - 2)\end{aligned}$$

The two factors are $(x - 3)$ and $(3x - 2)$.

Sum of the two factors $= (x - 3) + (3x - 2) = 4x - 5$

Hence, option 3 is the correct option.

Using the identity,

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

Factorizing,

$$\begin{aligned}\Rightarrow a^2 - 8a - 16b^2 + 16 \\[1em] \Rightarrow (a^2 - 8a + 16) - 16b^2 \\[1em] \Rightarrow (a - 4)^2 - (4b)^2 \\[1em] \Rightarrow [(a - 4) + 4b][(a - 4) - 4b] \\[1em] \Rightarrow (a - 4 + 4b)(a - 4 - 4b)\end{aligned}$$

The given expression is factorised into two factors, of which one is $(a - 4 - 4b)$.

Hence, option 2 is the correct option.

Area of the floor of the hall = length × breadth

So, we factorise the given expression to get the length and the breadth.

We want to find two integers whose sum is 9 and product is 20.

By trial, we see that 4 + 5 = 9 and 4 × 5 = 20.

Factorizing,

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

So, length = $(a + 5)$ m and breadth = $(a + 4)$ m

Substituting $a = 2$, we get

⇒ length = (2 + 5) m = 7 m

⇒ breadth = (2 + 4) m = 6 m

Hence, option 2 is the correct option.