Factorisation

Factorising,

⇒ $8xy^3 + 12x^2y^2$

HCF of $8xy^3$ and $12x^2y^2$:

HCF of numerical coefficients = HCF of 8 and 12 = 4

HCF of literal coefficients = HCF of $xy^3$ and $x^2y^2$

= product of each common literal raised to the lowest power

= $xy^2$

∴ HCF of $8xy^3$ and $12x^2y^2$ = 4 × $xy^2$ = $4xy^2$

Dividing each term by $4xy^2$ and writing $4xy^2$ outside the bracket,

⇒ $4xy^2(2y + 3x)$

Hence, $\bm{8xy^3 + 12x^2y^2 = 4xy^2(2y + 3x)}$

Factorising,

⇒ $15ax^3 - 9ax^2$

HCF of $15ax^3$ and $9ax^2$:

HCF of numerical coefficients = HCF of 15 and 9 = 3

HCF of literal coefficients = HCF of $ax^3$ and $ax^2$

= product of each common literal raised to the lowest power

= $ax^2$

∴ HCF of $15ax^3$ and $9ax^2$ = 3 × $ax^2$ = $3ax^2$

Dividing each term by $3ax^2$ and writing $3ax^2$ outside the bracket,

⇒ $3ax^2(5x - 3)$

Hence, $\bm{15ax^3 - 9ax^2 = 3ax^2(5x - 3)}$

Factorising,

⇒ $21py^2 - 56py$

HCF of $21py^2$ and $56py$:

HCF of numerical coefficients = HCF of 21 and 56 = 7

HCF of literal coefficients = HCF of $py^2$ and $py$

= product of each common literal raised to the lowest power

= $py$

∴ HCF of $21py^2$ and $56py$ = 7 × $py$ = $7py$

Dividing each term by $7py$ and writing $7py$ outside the bracket,

⇒ $7py(3y - 8)$

Hence, $\bm{21py^2 - 56py = 7py(3y - 8)}$

Factorising,

⇒ $4x^3 - 6x^2$

HCF of $4x^3$ and $6x^2$:

HCF of numerical coefficients = HCF of 4 and 6 = 2

HCF of literal coefficients = HCF of $x^3$ and $x^2$

= product of each common literal raised to the lowest power

= $x^2$

∴ HCF of $4x^3$ and $6x^2$ = 2 × $x^2$ = $2x^2$

Dividing each term by $2x^2$ and writing $2x^2$ outside the bracket,

⇒ $2x^2(2x - 3)$

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

Factorising,

⇒ $25abc^2 - 15a^2b^2c$

HCF of $25abc^2$ and $15a^2b^2c$:

HCF of numerical coefficients = HCF of 25 and 15 = 5

HCF of literal coefficients = HCF of $abc^2$ and $a^2b^2c$

= product of each common literal raised to the lowest power

= $abc$

∴ HCF of $25abc^2$ and $15a^2b^2c$ = 5 × $abc$ = $5abc$

Dividing each term by $5abc$ and writing $5abc$ outside the bracket,

⇒ $5abc(5c - 3ab)$

Hence, $\bm{25abc^2 - 15a^2b^2c = 5abc(5c - 3ab)}$

Factorising,

⇒ $x^2yz + xy^2z + xyz^2$

HCF of $x^2yz$, $xy^2z$ and $xyz^2$:

HCF of numerical coefficients = HCF of 1, 1 and 1 = 1

HCF of literal coefficients = HCF of $x^2yz$, $xy^2z$ and $xyz^2$

= product of each common literal raised to the lowest power

= $xyz$

∴ HCF of $x^2yz$, $xy^2z$ and $xyz^2$ = 1 × $xyz$ = $xyz$

Dividing each term by $xyz$ and writing $xyz$ outside the bracket,

⇒ $xyz(x + y + z)$

Hence, $\bm{x^2yz + xy^2z + xyz^2 = xyz(x + y + z)}$

Factorising,

⇒ $8x^3 - 6x^2 + 10x$

HCF of $8x^3$, $6x^2$ and $10x$:

HCF of numerical coefficients = HCF of 8, 6 and 10 = 2

HCF of literal coefficients = HCF of $x^3$, $x^2$ and $x$

= product of each common literal raised to the lowest power

= $x$

∴ HCF of $8x^3$, $6x^2$ and $10x$ = 2 × $x$ = $2x$

Dividing each term by $2x$ and writing $2x$ outside the bracket,

⇒ $2x(4x^2 - 3x + 5)$

Hence, $\bm{8x^3 - 6x^2 + 10x = 2x(4x^2 - 3x + 5)}$

Factorising,

⇒ $14mn + 22m - 62p$

HCF of $14mn$, $22m$ and $62p$:

HCF of numerical coefficients = HCF of 14, 22 and 62 = 2

HCF of literal coefficients = HCF of $mn$, $m$ and $p$

= product of each common literal raised to the lowest power

= 1, as no literal is common to all the three terms

∴ HCF of $14mn$, $22m$ and $62p$ = 2 × 1 = 2

Dividing each term by 2 and writing 2 outside the bracket,

⇒ $2(7mn + 11m - 31p)$

Hence, $\bm{14mn + 22m - 62p = 2(7mn + 11m - 31p)}$

Factorising,

⇒ $18p^2q^2 - 24pq^2 + 30p^2q$

HCF of $18p^2q^2$, $24pq^2$ and $30p^2q$:

HCF of numerical coefficients = HCF of 18, 24 and 30 = 6

HCF of literal coefficients = HCF of $p^2q^2$, $pq^2$ and $p^2q$

= product of each common literal raised to the lowest power

= $pq$

∴ HCF of $18p^2q^2$, $24pq^2$ and $30p^2q$ = 6 × $pq$ = $6pq$

Dividing each term by $6pq$ and writing $6pq$ outside the bracket,

⇒ $6pq(3pq - 4q + 5p)$

Hence, $\bm{18p^2q^2 - 24pq^2 + 30p^2q = 6pq(3pq - 4q + 5p)}$

Factorising,

⇒ $27a^3b^3 - 18a^2b^3 + 75a^3b^2$

HCF of $27a^3b^3$, $18a^2b^3$ and $75a^3b^2$:

HCF of numerical coefficients = HCF of 27, 18 and 75 = 3

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

= product of each common literal raised to the lowest power

= $a^2b^2$

∴ HCF of $27a^3b^3$, $18a^2b^3$ and $75a^3b^2$ = 3 × $a^2b^2$ = $3a^2b^2$

Dividing each term by $3a^2b^2$ and writing $3a^2b^2$ outside the bracket,

⇒ $3a^2b^2(9ab - 6b + 25a)$

Hence, $\bm{27a^3b^3 - 18a^2b^3 + 75a^3b^2 = 3a^2b^2(9ab - 6b + 25a)}$

Factorising,

⇒ $15a(2p - 3q) - 10b(2p - 3q)$

HCF of $15a(2p - 3q)$ and $10b(2p - 3q)$:

HCF of numerical coefficients = HCF of 15 and 10 = 5

HCF of the remaining factors = HCF of $a(2p - 3q)$ and $b(2p - 3q)$

= product of each common factor raised to the lowest power

= $(2p - 3q)$

∴ HCF of $15a(2p - 3q)$ and $10b(2p - 3q)$ = 5 × $(2p - 3q)$ = $5(2p - 3q)$

Dividing each term by $5(2p - 3q)$ and writing $5(2p - 3q)$ outside the bracket,

⇒ $5(2p - 3q)(3a - 2b)$

Hence, $\bm{15a(2p - 3q) - 10b(2p - 3q) = 5(2p - 3q)(3a - 2b)}$

Factorising,

⇒ $3a(x^2 + y^2) + 6b(x^2 + y^2)$

HCF of $3a(x^2 + y^2)$ and $6b(x^2 + y^2)$:

HCF of numerical coefficients = HCF of 3 and 6 = 3

HCF of the remaining factors = HCF of $a(x^2 + y^2)$ and $b(x^2 + y^2)$

= product of each common factor raised to the lowest power

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

∴ HCF of $3a(x^2 + y^2)$ and $6b(x^2 + y^2)$ = 3 × $(x^2 + y^2)$ = $3(x^2 + y^2)$

Dividing each term by $3(x^2 + y^2)$ and writing $3(x^2 + y^2)$ outside the bracket,

⇒ $3(x^2 + y^2)(a + 2b)$

Hence, $\bm{3a(x^2 + y^2) + 6b(x^2 + y^2) = 3(x^2 + y^2)(a + 2b)}$

Factorising,

⇒ $6(x + 2y)^3 + 8(x + 2y)^2$

HCF of $6(x + 2y)^3$ and $8(x + 2y)^2$:

HCF of numerical coefficients = HCF of 6 and 8 = 2

HCF of the remaining factors = HCF of $(x + 2y)^3$ and $(x + 2y)^2$

= product of each common factor raised to the lowest power

= $(x + 2y)^2$

∴ HCF of $6(x + 2y)^3$ and $8(x + 2y)^2$ = 2 × $(x + 2y)^2$ = $2(x + 2y)^2$

Dividing each term by $2(x + 2y)^2$ and writing $2(x + 2y)^2$ outside the bracket,

⇒ $2(x + 2y)^2(3(x + 2y) + 4)$

⇒ $2(x + 2y)^2(3x + 6y + 4)$

Hence, $\bm{6(x + 2y)^3 + 8(x + 2y)^2 = 2(x + 2y)^2(3x + 6y + 4)}$

Factorising,

⇒ $14(a - 3b)^3 - 21p(a - 3b)$

HCF of $14(a - 3b)^3$ and $21p(a - 3b)$:

HCF of numerical coefficients = HCF of 14 and 21 = 7

HCF of the remaining factors = HCF of $(a - 3b)^3$ and $p(a - 3b)$

= product of each common factor raised to the lowest power

= $(a - 3b)$

∴ HCF of $14(a - 3b)^3$ and $21p(a - 3b)$ = 7 × $(a - 3b)$ = $7(a - 3b)$

Dividing each term by $7(a - 3b)$ and writing $7(a - 3b)$ outside the bracket,

⇒ $7(a - 3b)[2(a - 3b)^2 - 3p]$

Hence, $\bm{14(a - 3b)^3 - 21p(a - 3b) = 7(a - 3b)[2(a - 3b)^2 - 3p]}$

Factorising,

⇒ $10a(2p + q)^3 - 15b(2p + q)^2 + 35(2p + q)$

HCF of $10a(2p + q)^3$, $15b(2p + q)^2$ and $35(2p + q)$:

HCF of numerical coefficients = HCF of 10, 15 and 35 = 5

HCF of the remaining factors = HCF of $a(2p + q)^3$, $b(2p + q)^2$ and $(2p + q)$

= product of each common factor raised to the lowest power

= $(2p + q)$

∴ HCF of $10a(2p + q)^3$, $15b(2p + q)^2$ and $35(2p + q)$ = 5 × $(2p + q)$ = $5(2p + q)$

Dividing each term by $5(2p + q)$ and writing $5(2p + q)$ outside the bracket,

⇒ $5(2p + q)[2a(2p + q)^2 - 3b(2p + q) + 7]$

Hence, $\bm{10a(2p + q)^3 - 15b(2p + q)^2 + 35(2p + q) = 5(2p + q)[2a(2p + q)^2 - 3b(2p + q) + 7]}$

Factorising,

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

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

Factorising,

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

Hence, $\bm{y^2 - yz - 5y + 5z = (y - z)(y - 5)}$

Factorising,

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

Hence, $\bm{5xy + 7y - 5y^2 - 7x = (x - y)(5y - 7)}$

Factorising,

$$\begin{aligned}\Rightarrow 5p^2 - 8pq - 10p + 16q \\[1em] \Rightarrow (5p^2 - 8pq) - (10p - 16q) \\[1em] \Rightarrow p(5p - 8q) - 2(5p - 8q) \\[1em] \Rightarrow (5p - 8q)(p - 2).\end{aligned}$$

Hence, $\bm{5p^2 - 8pq - 10p + 16q = (5p - 8q)(p - 2)}$

Factorising,

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

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

Factorising,

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

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

Factorising,

$$\begin{aligned}\Rightarrow 6xy^2 - 3xy - 10y + 5 \\[1em] \Rightarrow (6xy^2 - 3xy) - (10y - 5) \\[1em] \Rightarrow 3xy(2y - 1) - 5(2y - 1) \\[1em] \Rightarrow (2y - 1)(3xy - 5).\end{aligned}$$

Hence, $\bm{6xy^2 - 3xy - 10y + 5 = (2y - 1)(3xy - 5)}$

As $3ax$ and $6ay$ are to be grouped together, we interchange the positions of $-8by$ and $4bx$

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

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

Factorising,

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

Hence, $\bm{x^2 + xy(1 + y) + y^3 = (x + y)(x + y^2)}$

Factorising,

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

Hence, $\bm{y^2 - xy(1 - x) - x^3 = (y - x)(y + x^2)}$

Factorising,

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

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

Factorising,

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

Hence, $\bm{2a - 4b - xa + 2bx = (a - 2b)(2 - x)}$

Factorising,

$$\begin{aligned}\Rightarrow 5ph - 10qk + 2rph - 4qrk \\[1em] \Rightarrow (5ph - 10qk) + (2rph - 4qrk) \\[1em] \Rightarrow 5(ph - 2qk) + 2r(ph - 2qk) \\[1em] \Rightarrow (ph - 2qk)(5 + 2r).\end{aligned}$$

Hence, $\bm{5ph - 10qk + 2rph - 4qrk = (5 + 2r)(ph - 2qk)}$

Factorising,

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

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

Factorising,

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

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 [∵ $(x - y)^2 = x^2 - 2xy + y^2$]

Factorising,

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

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

Factorising,

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

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

Factorising,

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

Hence, $\bm{3x^2y - 3xy + 12x - 12 = 3(x - 1)(xy + 4)}$

Factorising,

$$\begin{aligned}\Rightarrow a^2b + ab^2 - abc - b^2c + axy + bxy \\[1em] \Rightarrow (a^2b + ab^2) - (abc + b^2c) + (axy + bxy) \\[1em] \Rightarrow ab(a + b) - bc(a + b) + xy(a + b) \\[1em] \Rightarrow (a + b)(ab - bc + xy).\end{aligned}$$

Hence, $\bm{a^2b + ab^2 - abc - b^2c + axy + bxy = (a + b)(ab - bc + xy)}$

Factorising,

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

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

Factorising,

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

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

Factorising,

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

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

Factorising,

⇒ $x^2 - 12x + 36$

⇒ $(x)^2 - 2 \times x \times 6 + (6)^2$

Using the identity,

[∵ $(x - y)^2 = x^2 - 2xy + y^2$]

⇒ $(x - 6)^2$

⇒ $(x - 6)(x - 6)$

Hence, $\bm{x^2 - 12x + 36 = (x - 6)(x - 6) = (x - 6)^2}$

Factorising,

⇒ $36p^2 - 60pq + 25q^2$

⇒ $(6p)^2 - 2 \times 6p \times 5q + (5q)^2$

Using the identity,

[∵ $(x - y)^2 = x^2 - 2xy + y^2$]

⇒ $(6p - 5q)^2$

⇒ $(6p - 5q)(6p - 5q)$

Hence, $\bm{36p^2 - 60pq + 25q^2 = (6p - 5q)(6p - 5q) = (6p - 5q)^2}$

Factorising,

⇒ $9x^2 + 66xy + 121y^2$

⇒ $(3x)^2 + 2 \times 3x \times 11y + (11y)^2$

Using the identity,

[∵ $(x + y)^2 = x^2 + 2xy + y^2$]

⇒ $(3x + 11y)^2$

⇒ $(3x + 11y)(3x + 11y)$

Hence, $\bm{9x^2 + 66xy + 121y^2 = (3x + 11y)(3x + 11y) = (3x + 11y)^2}$

Factorising,

⇒ $a^4 + 6a^2b^2 + 9b^4$

⇒ $(a^2)^2 + 2 \times a^2 \times 3b^2 + (3b^2)^2$

Using the identity,

[∵ $(x + y)^2 = x^2 + 2xy + y^2$]

⇒ $(a^2 + 3b^2)^2$

⇒ $(a^2 + 3b^2)(a^2 + 3b^2)$

Hence, $\bm{a^4 + 6a^2b^2 + 9b^4 = (a^2 + 3b^2)(a^2 + 3b^2) = (a^2 + 3b^2)^2}$

Using the identity,

[∵ $(x + y)^2 = x^2 + 2xy + y^2$]

Factorising,

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

Hence, $\bm{x^2 + \frac{1}{x^2} + 2 = \Big(x + \frac{1}{x}\Big)\Big(x + \frac{1}{x}\Big) = \Big(x + \frac{1}{x}\Big)^2}$

Using the identity,

[∵ $(x + y)^2 = x^2 + 2xy + y^2$]

Factorising,

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

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

Factorising,

⇒ $4p^2 - 9$

⇒ $(2p)^2 - (3)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(2p + 3)(2p - 3)$

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

Factorising,

⇒ $4x^2 - 169y^2$

⇒ $(2x)^2 - (13y)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(2x + 13y)(2x - 13y)$

Hence, $\bm{4x^2 - 169y^2 = (2x + 13y)(2x - 13y)}$

Factorising,

⇒ $9x^2y^2 - 25$

⇒ $(3xy)^2 - (5)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(3xy + 5)(3xy - 5)$

Hence, $\bm{9x^2y^2 - 25 = (3xy + 5)(3xy - 5)}$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

Factorising,

$$\begin{aligned}\Rightarrow 16x^2 - \frac{1}{144}\\[1em] \Rightarrow (4x)^2 - \Big(\frac{1}{12}\Big)^2\\[1em] \Rightarrow \Big(4x + \frac{1}{12}\Big)\Big(4x - \frac{1}{12}\Big)\end{aligned}$$

Hence, $\bm{16x^2 - \frac{1}{144} = \Big(4x + \frac{1}{12}\Big)\Big(4x - \frac{1}{12}\Big)}$

Factorising,

⇒ $20x^2 - 45y^2$

⇒ $5(4x^2 - 9y^2)$

⇒ $5((2x)^2 - (3y)^2)$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $5(2x + 3y)(2x - 3y)$

Hence, $\bm{20x^2 - 45y^2 = 5(2x + 3y)(2x - 3y)}$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

Factorising,

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

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

Factorising,

⇒ $(2a + 3b)^2 - 16c^2$

⇒ $(2a + 3b)^2 - (4c)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $((2a + 3b) + (4c))((2a + 3b) - (4c))$

⇒ $(2a + 3b + 4c)(2a + 3b - 4c)$

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

Factorising,

⇒ $1 - (b - c)^2$

⇒ $(1)^2 - (b - c)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(1 + (b - c))(1 - (b - c))$

⇒ $(1 + b - c)(1 - b + c)$

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

Factorising,

⇒ $9(x + y)^2 - x^2$

⇒ $(3(x + y))^2 - (x)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(3(x + y) + x)(3(x + y) - x)$

⇒ $(3x + 3y + x)(3x + 3y - x)$

⇒ $(4x + 3y)(2x + 3y)$

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

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

Factorising,

⇒ $(2m + 3n)^2 - (3m + 2n)^2$

⇒ $((2m + 3n) + (3m + 2n))((2m + 3n) - (3m + 2n))$

⇒ $(2m + 3n + 3m + 2n)(2m + 3n - 3m - 2n)$

⇒ $(5m + 5n)(n - m)$

⇒ $5(m + n)(n - m)$

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

Factorising,

⇒ $25(a + b)^2 - 16(a - b)^2$

⇒ $(5(a + b))^2 - (4(a - b))^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(5(a + b) + 4(a - b))(5(a + b) - 4(a - b))$

⇒ $(5a + 5b + 4a - 4b)(5a + 5b - 4a + 4b)$

⇒ $(9a + b)(a + 9b)$

Hence, $\bm{25(a + b)^2 - 16(a - b)^2 = (9a + b)(a + 9b)}$

Factorising,

⇒ $9(3x + 2)^2 - 4(2x - 1)^2$

⇒ $(3(3x + 2))^2 - (2(2x - 1))^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(3(3x + 2) + 2(2x - 1))(3(3x + 2) - 2(2x - 1))$

⇒ $(9x + 6 + 4x - 2)(9x + 6 - 4x + 2)$

⇒ $(13x + 4)(5x + 8)$

Hence, $\bm{9(3x + 2)^2 - 4(2x - 1)^2 = (13x + 4)(5x + 8)}$

Factorising,

⇒ $x^3 - 25x$

⇒ $x(x^2 - 25)$

⇒ $x((x)^2 - (5)^2)$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $x(x + 5)(x - 5)$

Hence, $\bm{x^3 - 25x = x(x + 5)(x - 5)}$

Factorising,

⇒ $63p^2q^2 - 7$

⇒ $7(9p^2q^2 - 1)$

⇒ $7((3pq)^2 - (1)^2)$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $7(3pq + 1)(3pq - 1)$

Hence, $\bm{63p^2q^2 - 7 = 7(3pq + 1)(3pq - 1)}$

Factorising,

⇒ $32a^2b - 72b^3$

⇒ $8b(4a^2 - 9b^2)$

⇒ $8b((2a)^2 - (3b)^2)$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $8b(2a + 3b)(2a - 3b)$

Hence, $\bm{32a^2b - 72b^3 = 8b(2a + 3b)(2a - 3b)}$

Factorising,

⇒ $9(a + b)^3 - 25(a + b)$

⇒ $(a + b)(9(a + b)^2 - 25)$

⇒ $(a + b)((3(a + b))^2 - (5)^2)$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(a + b)(3(a + b) + 5)(3(a + b) - 5)$

⇒ $(a + b)(3a + 3b + 5)(3a + 3b - 5)$

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

Factorising,

⇒ $x^2 - y^2 - 2y - 1$

⇒ $x^2 - (y^2 + 2y + 1)$

Using the identity,

[∵ $(x + y)^2 = x^2 + 2xy + y^2$]

⇒ $(x)^2 - (y + 1)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(x + (y + 1))(x - (y + 1))$

⇒ $(x + y + 1)(x - y - 1)$

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

Factorising,

⇒ $p^2 - 4pq + 4q^2 - r^2$

⇒ $(p^2 - 4pq + 4q^2) - r^2$

Using the identity,

[∵ $(x - y)^2 = x^2 - 2xy + y^2$]

⇒ $(p - 2q)^2 - (r)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $((p - 2q) + r)((p - 2q) - r)$

⇒ $(p - 2q + r)(p - 2q - r)$

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

Factorising,

⇒ $9x^2 - y^2 + 4y - 4$

⇒ $9x^2 - (y^2 - 4y + 4)$

Using the identity,

[∵ $(x - y)^2 = x^2 - 2xy + y^2$]

⇒ $(3x)^2 - (y - 2)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(3x + (y - 2))(3x - (y - 2))$

⇒ $(3x + y - 2)(3x - y + 2)$

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

Factorising,

⇒ $4a^2 - 4b^2 + 4a + 1$

⇒ $(4a^2 + 4a + 1) - 4b^2$

⇒ $[(2a)^2 + 2 \times 2a \times 1 + 1] - (4b^2)$

Using the identity,

[∵ $(x + y)^2 = x^2 + 2xy + y^2$]

⇒ $(2a + 1)^2 - (2b)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $((2a + 1) + 2b)((2a + 1) - 2b)$

⇒ $(2a + 2b + 1)(2a - 2b + 1)$

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

Factorising,

⇒ $625 - p^4$

⇒ $(25)^2 - (p^2)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(25 + p^2)(25 - p^2)$

⇒ $(25 + p^2)((5)^2 - (p)^2)$

⇒ $(25 + p^2)(5 + p)(5 - p)$

Hence, $\bm{625 - p^4 = (25 + p^2)(5 + p)(5 - p)}$

Factorising,

⇒ $5y^5 - 405y$

⇒ $5y(y^4 - 81)$

⇒ $5y((y^2)^2 - (9)^2)$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $5y(y^2 + 9)(y^2 - 9)$

⇒ $5y(y^2 + 9)((y)^2 - (3)^2)$

⇒ $5y(y^2 + 9)(y + 3)(y - 3)$

Hence, $\bm{5y^5 - 405y = 5y(y^2 + 9)(y + 3)(y - 3)}$

Factorising,

⇒ $x^4 - y^4 + x^2 - y^2$

⇒ $(x^4 - y^4) + (x^2 - y^2)$

⇒ $((x^2)^2 - (y^2)^2) + (x^2 - y^2)$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(x^2 + y^2)(x^2 - y^2) + 1(x^2 - y^2)$

⇒ $(x^2 - y^2)(x^2 + y^2 + 1)$

⇒ $(x + y)(x - y)(x^2 + y^2 + 1)$

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

Factorising,

⇒ $64a^2 - 9b^2 + 42bc - 49c^2$

⇒ $64a^2 - (9b^2 - 42bc + 49c^2)$

Using the identity,

[∵ $(x - y)^2 = x^2 - 2xy + y^2$]

⇒ $(8a)^2 - (3b - 7c)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(8a + (3b - 7c))(8a - (3b - 7c))$

⇒ $(8a + 3b - 7c)(8a - 3b + 7c)$

Hence, $\bm{64a^2 - 9b^2 + 42bc - 49c^2 = (8a + 3b - 7c)(8a - 3b + 7c)}$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

Evaluating,

$$\begin{aligned}\Rightarrow \frac{38^2 - 22^2}{16}\\[1em] \Rightarrow \frac{(38 + 22)(38 - 22)}{16}\\[1em] \Rightarrow \frac{60 \times 16}{16}\\[1em] \Rightarrow \frac{60 \times \cancel{16}}{\cancel{16}}\\[1em] \Rightarrow 60\end{aligned}$$

Hence, $\bm{\frac{38^2 - 22^2}{16} = 60}$

Using the identities,

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

Evaluating,

$$\begin{aligned}\Rightarrow \frac{(5.3)^2 - (3.3)^2}{(11.2)^2 - (11.2) \times (2.4) + (1.2)^2}\\[1em] \Rightarrow \frac{(5.3)^2 - (3.3)^2}{(11.2)^2 - 2 \times (11.2) \times (1.2) + (1.2)^2}\\[1em] \Rightarrow \frac{(5.3 + 3.3)(5.3 - 3.3)}{(11.2 - 1.2)^2}\\[1em] \Rightarrow \frac{8.6 \times 2}{(10)^2}\\[1em] \Rightarrow \frac{17.2}{100}\\[1em] \Rightarrow 0.172\end{aligned}$$

Hence, $\bm{\frac{(5.3)^2 - (3.3)^2}{(11.2)^2 - (11.2) \times (2.4) + (1.2)^2} = 0.172}$

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

By trial, we see that 1 + 2 = 3 and 1 × 2 = 2

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

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

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

By trial, we see that 4 + 6 = 10 and 4 × 6 = 24

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

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

We want to find two integers whose sum is −7 and product is 12.

By trial, we see that (−3) + (−4) = −7 and (−3) × (−4) = 12

$$\begin{aligned}\Rightarrow y^2 - 7y + 12 \\[1em] \Rightarrow y^2 - (3 + 4)y + 12 \\[1em] \Rightarrow y^2 - 3y - 4y + 12 \\[1em] \Rightarrow y(y - 3) - 4(y - 3) \\[1em] \Rightarrow (y - 3)(y - 4).\end{aligned}$$

Hence, $\bm{y^2 - 7y + 12 = (y - 3)(y - 4)}$

We want to find two integers whose sum is −23 and product is 42.

By trial, we see that (−2) + (−21) = −23 and (−2) × (−21) = 42

$$\begin{aligned}\Rightarrow m^2 - 23m + 42 \\[1em] \Rightarrow m^2 - (2 + 21)m + 42 \\[1em] \Rightarrow m^2 - 2m - 21m + 42 \\[1em] \Rightarrow m(m - 2) - 21(m - 2) \\[1em] \Rightarrow (m - 2)(m - 21).\end{aligned}$$

Hence, $\bm{m^2 - 23m + 42 = (m - 2)(m - 21)}$

We want to find two integers whose sum is −5 and product is −24.

By trial, we see that 3 + (−8) = −5 and 3 × (−8) = −24

$$\begin{aligned}\Rightarrow y^2 - 5y - 24 \\[1em] \Rightarrow y^2 + (3 - 8)y - 24 \\[1em] \Rightarrow y^2 + 3y - 8y - 24 \\[1em] \Rightarrow y(y + 3) - 8(y + 3) \\[1em] \Rightarrow (y + 3)(y - 8).\end{aligned}$$

Hence, $\bm{y^2 - 5y - 24 = (y + 3)(y - 8)}$

We want to find two integers whose sum is 23 and product is −108.

By trial, we see that 27 + (−4) = 23 and 27 × (−4) = −108

$$\begin{aligned}\Rightarrow t^2 + 23t - 108 \\[1em] \Rightarrow t^2 + (27 - 4)t - 108 \\[1em] \Rightarrow t^2 + 27t - 4t - 108 \\[1em] \Rightarrow t(t + 27) - 4(t + 27) \\[1em] \Rightarrow (t + 27)(t - 4).\end{aligned}$$

Hence, $\bm{t^2 + 23t - 108 = (t + 27)(t - 4)}$

We want to find two integers whose sum is 14 and product is 3 × 8 i.e. 24.

By trial, we see that 12 + 2 = 14 and 12 × 2 = 24

$$\begin{aligned}\Rightarrow 3x^2 + 14x + 8 \\[1em] \Rightarrow 3x^2 + 12x + 2x + 8 \\[1em] \Rightarrow 3x(x + 4) + 2(x + 4) \\[1em] \Rightarrow (x + 4)(3x + 2).\end{aligned}$$

Hence, $\bm{3x^2 + 14x + 8 = (3x + 2)(x + 4)}$

We want to find two integers whose sum is 10 and product is 3 × 8 i.e. 24.

By trial, we see that 6 + 4 = 10 and 6 × 4 = 24

$$\begin{aligned}\Rightarrow 3y^2 + 10y + 8 \\[1em] \Rightarrow 3y^2 + 6y + 4y + 8 \\[1em] \Rightarrow 3y(y + 2) + 4(y + 2) \\[1em] \Rightarrow (y + 2)(3y + 4).\end{aligned}$$

Hence, $\bm{3y^2 + 10y + 8 = (3y + 4)(y + 2)}$

We want to find two integers whose sum is −23 and product is 14 × 8 i.e. 112.

By trial, we see that (−16) + (−7) = −23 and (−16) × (−7) = 112

$$\begin{aligned}\Rightarrow 14x^2 - 23x + 8 \\[1em] \Rightarrow 14x^2 - 16x - 7x + 8 \\[1em] \Rightarrow 2x(7x - 8) - 1(7x - 8) \\[1em] \Rightarrow (7x - 8)(2x - 1).\end{aligned}$$

Hence, $\bm{14x^2 - 23x + 8 = (7x - 8)(2x - 1)}$

We want to find two integers whose sum is −1 and product is 12 × (−35) i.e. −420.

By trial, we see that 20 + (−21) = −1 and 20 × (−21) = −420

$$\begin{aligned}\Rightarrow 12x^2 - x - 35 \\[1em] \Rightarrow 12x^2 + 20x - 21x - 35 \\[1em] \Rightarrow 4x(3x + 5) - 7(3x + 5) \\[1em] \Rightarrow (3x + 5)(4x - 7).\end{aligned}$$

Hence, $\bm{12x^2 - x - 35 = (4x - 7)(3x + 5)}$

We want to find two integers whose sum is 11 and product is 6 × (−10) i.e. −60.

By trial, we see that 15 + (−4) = 11 and 15 × (−4) = −60

$$\begin{aligned}\Rightarrow 6x^2 + 11x - 10 \\[1em] \Rightarrow 6x^2 + 15x - 4x - 10 \\[1em] \Rightarrow 3x(2x + 5) - 2(2x + 5) \\[1em] \Rightarrow (2x + 5)(3x - 2).\end{aligned}$$

Hence, $\bm{6x^2 + 11x - 10 = (2x + 5)(3x - 2)}$

We want to find two integers whose sum is −4 and product is (−12) × 5 i.e. −60.

By trial, we see that (−10) + 6 = −4 and (−10) × 6 = −60

$$\begin{aligned}\Rightarrow 5 - 4x - 12x^2 \\[1em] \Rightarrow 5 - 10x + 6x - 12x^2 \\[1em] \Rightarrow 5(1 - 2x) + 6x(1 - 2x) \\[1em] \Rightarrow (1 - 2x)(5 + 6x).\end{aligned}$$

Hence, $\bm{5 - 4x - 12x^2 = (1 - 2x)(5 + 6x)}$

We want to find two integers whose sum is −18 and product is (−63) × 1 i.e. −63.

By trial, we see that (−21) + 3 = −18 and (−21) × 3 = −63

$$\begin{aligned}\Rightarrow 1 - 18y - 63y^2 \\[1em] \Rightarrow 1 - 21y + 3y - 63y^2 \\[1em] \Rightarrow 1(1 - 21y) + 3y(1 - 21y) \\[1em] \Rightarrow (1 - 21y)(1 + 3y).\end{aligned}$$

Hence, $\bm{1 - 18y - 63y^2 = (1 - 21y)(1 + 3y)}$

We want to find two integers whose sum is −5 and product is 3 × (−12) i.e. −36.

By trial, we see that (−9) + 4 = −5 and (−9) × 4 = −36

$$\begin{aligned}\Rightarrow 3x^2 - 5xy - 12y^2 \\[1em] \Rightarrow 3x^2 - 9xy + 4xy - 12y^2 \\[1em] \Rightarrow 3x(x - 3y) + 4y(x - 3y) \\[1em] \Rightarrow (x - 3y)(3x + 4y).\end{aligned}$$

Hence, $\bm{3x^2 - 5xy - 12y^2 = (3x + 4y)(x - 3y)}$

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

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

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

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

We want to find two integers whose sum is −21 and product is 10 × 9 i.e. 90.

By trial, we see that (−15) + (−6) = −21 and (−15) × (−6) = 90

$$\begin{aligned}\Rightarrow 10p^2q^2 - 21pq + 9 \\[1em] \Rightarrow 10p^2q^2 - 15pq - 6pq + 9 \\[1em] \Rightarrow 5pq(2pq - 3) - 3(2pq - 3) \\[1em] \Rightarrow (2pq - 3)(5pq - 3).\end{aligned}$$

Hence, $\bm{10p^2q^2 - 21pq + 9 = (2pq - 3)(5pq - 3)}$

We want to find two integers whose sum is 1 and product is 2 × (−45) i.e. −90.

By trial, we see that 10 + (−9) = 1 and 10 × (−9) = −90

$$\begin{aligned}\Rightarrow 2a^2b^2 + ab - 45 \\[1em] \Rightarrow 2a^2b^2 + 10ab - 9ab - 45 \\[1em] \Rightarrow 2ab(ab + 5) - 9(ab + 5) \\[1em] \Rightarrow (ab + 5)(2ab - 9).\end{aligned}$$

Hence, $\bm{2a^2b^2 + ab - 45 = (ab + 5)(2ab - 9)}$

Solving,

$$\begin{aligned}\Rightarrow x(12x + 7) - 10 \\[1em] \Rightarrow 12x^2 + 7x - 10\end{aligned}$$

We want to find two integers whose sum is 7 and product is 12 × (−10) i.e. −120.

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

$$\begin{aligned}\Rightarrow 12x^2 + 15x - 8x - 10 \\[1em] \Rightarrow 3x(4x + 5) - 2(4x + 5) \\[1em] \Rightarrow (4x + 5)(3x - 2).\end{aligned}$$

Hence, $\bm{x(12x + 7) - 10 = (4x + 5)(3x - 2)}$

Let $a + b = m$, then

$(a + b)^2 - 11(a + b) - 42 = m^2 - 11m - 42$

We want to find two integers whose sum is −11 and product is −42.

By trial, we see that 3 + (−14) = −11 and 3 × (−14) = −42

$$\begin{aligned}\Rightarrow m^2 - 11m - 42 \\[1em] \Rightarrow m^2 + 3m - 14m - 42 \\[1em] \Rightarrow m(m + 3) - 14(m + 3) \\[1em] \Rightarrow (m + 3)(m - 14).\end{aligned}$$

Substituting $m = a + b$, we get

$(a + b)^2 - 11(a + b) - 42 = (a + b + 3)(a + b - 14)$

Hence, $\bm{(a + b)^2 - 11(a + b) - 42 = (a + b + 3)(a + b - 14)}$

Let $p + q = m$, then

$8 + 6(p + q) - 5(p + q)^2 = 8 + 6m - 5m^2$

We want to find two integers whose sum is 6 and product is (−5) × 8 i.e. −40.

By trial, we see that 10 + (−4) = 6 and 10 × (−4) = −40

$$\begin{aligned}\Rightarrow 8 + 6m - 5m^2 \\[1em] \Rightarrow 8 + 10m - 4m - 5m^2 \\[1em] \Rightarrow 2(4 + 5m) - m(4 + 5m) \\[1em] \Rightarrow (4 + 5m)(2 - m).\end{aligned}$$

Substituting $m = p + q$, we get

$8 + 6(p + q) - 5(p + q)^2 = (4 + 5p + 5q)(2 - p - q)$

Hence, $\bm{8 + 6(p + q) - 5(p + q)^2 = (4 + 5p + 5q)(2 - p - q)}$

Let $x - 2y = m$, then

$(x - 2y)^2 - 6(x - 2y) + 5 = m^2 - 6m + 5$

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

By trial, we see that (−1) + (−5) = −6 and (−1) × (−5) = 5

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

Substituting $m = x - 2y$, we get

$(x - 2y)^2 - 6(x - 2y) + 5 = (x - 2y - 1)(x - 2y - 5)$

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

Let $2x - 3y = m$, then

$7 + 10(2x - 3y) - 8(2x - 3y)^2 = 7 + 10m - 8m^2$

We want to find two integers whose sum is 10 and product is (−8) × 7 i.e. −56.

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

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

Substituting $m = 2x - 3y$, we get

$7 + 10(2x - 3y) - 8(2x - 3y)^2 = (1 + 4x - 6y)(7 - 8x + 12y)$

Hence, $\bm{7 + 10(2x - 3y) - 8(2x - 3y)^2 = (1 + 4x - 6y)(7 - 8x + 12y)}$

Factorising the numerator $35x + 28$, we have

⇒ $(35x + 28)$

⇒ $7(5x + 4)$

Solving,

$$\begin{aligned}\Rightarrow \frac{35x + 28}{5x + 4}\\[1em] \Rightarrow \frac{7(5x + 4)}{(5x + 4)}\\[1em] \Rightarrow \frac{7\cancel{(5x + 4)}}{\cancel{(5x + 4)}}\\[1em] \Rightarrow 7\end{aligned}$$

Hence, $\bm{(35x + 28) \div (5x + 4) = 7}$

Factorising $9r - 27$, we get $9(r - 3)$

Solving,

$$\begin{aligned}\Rightarrow \frac{7p^2q^2(9r - 27)}{63pq(r - 3)}\\[1em] \Rightarrow \frac{7p^2q^2 \times 9(r - 3)}{63pq(r - 3)}\\[1em] \Rightarrow \frac{63p^2q^2(r - 3)}{63pq(r - 3)}\\[1em] \Rightarrow \frac{\cancel{63}p^2q^2\cancel{(r - 3)}}{\cancel{63}pq\cancel{(r - 3)}}\\[1em] \Rightarrow \frac{p^2q^2}{pq}\\[1em] \Rightarrow pq\end{aligned}$$

Hence, $\bm{7p^2q^2(9r - 27) \div 63pq(r - 3) = pq}$

Solving,

$$\begin{aligned}\Rightarrow \frac{6(2x + 7)(5x - 3)}{3(5x - 3)}\\[1em] \Rightarrow \frac{6(2x + 7)\cancel{(5x - 3)}}{3\cancel{(5x - 3)}}\\[1em] \Rightarrow \frac{6(2x + 7)}{3}\\[1em] \Rightarrow 2(2x + 7)\end{aligned}$$

Hence, $\bm{6(2x + 7)(5x - 3) \div 3(5x - 3) = 2(2x + 7)}$

Solving,

$$\begin{aligned}\Rightarrow \frac{33pq(p + 3)(2q - 5)}{11p(2q - 5)}\\[1em] \Rightarrow \frac{33\cancel{p}q(p + 3)\cancel{(2q - 5)}}{11\cancel{p}\cancel{(2q - 5)}}\\[1em] \Rightarrow 3q(p + 3)\end{aligned}$$

Hence, $\bm{33pq(p + 3)(2q - 5) \div 11p(2q - 5) = 3q(p + 3)}$

Factorising the numerator $7x^3 - 63x$, we get

⇒ $7x^3 - 63x$

⇒ $7x(x^2 - 9)$

⇒ $7x((x)^2 - (3)^2)$

⇒ $7x(x + 3)(x - 3)$

Solving,

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

Hence, $\bm{(7x^3 - 63x) \div 7(x - 3) = x(x + 3)}$

Factorising the numerator $3p^2 + 17p + 10$, we want two integers whose sum is 17 and product is 3 × 10 i.e. 30.

By trial, we see that 15 + 2 = 17 and 15 × 2 = 30

⇒ $3p^2 + 17p + 10$

⇒ $3p^2 + 15p + 2p + 10$

⇒ $3p(p + 5) + 2(p + 5)$

⇒ $(p + 5)(3p + 2)$

Solving,

$$\begin{aligned}\Rightarrow \frac{3p^2 + 17p + 10}{p + 5}\\[1em] \Rightarrow \frac{(p + 5)(3p + 2)}{(p + 5)}\\[1em] \Rightarrow \frac{\cancel{(p + 5)}(3p + 2)}{\cancel{(p + 5)}}\\[1em] \Rightarrow 3p + 2\end{aligned}$$

Hence, $\bm{(3p^2 + 17p + 10) \div (p + 5) = 3p + 2}$

Factorising $14y^2 + 43y - 21$, we want two integers whose sum is 43 and product is 14 × (−21) i.e. −294.

By trial, we see that 49 + (−6) = 43 and 49 × (−6) = −294

⇒ $14y^2 + 43y - 21$

⇒ $14y^2 + 49y - 6y - 21$

⇒ $7y(2y + 7) - 3(2y + 7)$

⇒ $(2y + 7)(7y - 3)$

Solving,

$$\begin{aligned}\Rightarrow \frac{10xy(14y^2 + 43y - 21)}{5x(7y - 3)}\\[1em] \Rightarrow \frac{10xy(2y + 7)(7y - 3)}{5x(7y - 3)}\\[1em] \Rightarrow \frac{10\cancel{x}y(2y + 7)\cancel{(7y - 3)}}{5\cancel{x}\cancel{(7y - 3)}}\\[1em] \Rightarrow \frac{10y(2y + 7)}{5}\\[1em] \Rightarrow 2y(2y + 7)\end{aligned}$$

Hence, $\bm{10xy(14y^2 + 43y - 21) \div 5x(7y - 3) = 2y(2y + 7)}$

Factorising $6p^2 - 13pq + 6q^2$, we want two integers whose sum is −13 and product is 6 × 6 i.e. 36.

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

⇒ $6p^2 - 13pq + 6q^2$

⇒ $6p^2 - 9pq - 4pq + 6q^2$

⇒ $3p(2p - 3q) - 2q(2p - 3q)$

⇒ $(2p - 3q)(3p - 2q)$

Solving,

$$\begin{aligned}\Rightarrow \frac{12pqr(6p^2 - 13pq + 6q^2)}{6pq(2p - 3q)}\\[1em] \Rightarrow \frac{12pqr(2p - 3q)(3p - 2q)}{6pq(2p - 3q)}\\[1em] \Rightarrow \frac{12\cancel{pq}r\cancel{(2p - 3q)}(3p - 2q)}{6\cancel{pq}\cancel{(2p - 3q)}}\\[1em] \Rightarrow \frac{12r(3p - 2q)}{6}\\[1em] \Rightarrow 2r(3p - 2q)\end{aligned}$$

Hence, $\bm{12pqr(6p^2 - 13pq + 6q^2) \div 6pq(2p - 3q) = 2r(3p - 2q)}$

(i) When an algebraic expression can be written as the product of two or more expressions then each of these expressions is called factor of the given expression.

(ii) The process of finding two or more expressions whose product is the given expression is called factorisation.

(iii) HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

(iv) HCF of literal coefficients = product of each common literal raised to the lowest power.

(v) To factorise the trinomial of the form $x^2 + px + q$, we need to find two integers $a$ and $b$ such that $a + b$ = $\bm{p}$ and $ab$ = $\bm{q}$.

(vi) To factorise the trinomial of the form $ax^2 + bx + c$, where $a$, $b$, and $c$ are integers, we split $b$ into two parts such that algebraic sum of these parts is $b$ and their product is $ac$

(i) Factorisation is the reverse process of multiplication.

True.

(ii) HCF of two or more polynomials (with integral coefficients) is the smallest common factor of the given polynomials.

False.

Reason — HCF of two or more polynomials (with integral coefficients) is the largest common factor of the given polynomials.

(iii) HCF of $6x^2y^2$ and $8xy^3$ is $2xy^2$

True.

(iv) Factorisation by grouping is possible only if the given polynomial contains an even number of terms.

True.

(v) To factorise the trinomial of the form $ax^2 + bx + c$ where, $a$, $b$, $c$ are integers we want to find two integers $A$ and $B$ such that $A + B = ac$ and $AB = b$

False.

Reason — We want to find two integers $A$ and $B$ such that $A + B = b$ and $AB = ac$.

(vi) Factors of $4x^2 - 12x + 9$ are $(2x - 3)(2x - 3)$

True.

(vii) The value of $p$ for $(51)^2 - (49)^2 = 100p$ is 2

True.

(viii) The difference of the squares of two consecutive natural numbers is their sum.

True.

HCF of numerical coefficients = HCF of 6, 24 and 12 = 6

HCF of literal coefficients = HCF of $abc$, $ab^2$ and $a^2b$

= product of each common literal raised to the lowest power

= ab

HCF of the given monomials = 6 × $ab$ = $6ab$

Hence, option 1 is the correct option.

HCF of $12a^2b$ and $15ab^2$ is $3ab$

Divide each term by $3ab$ and keep $3ab$ outside the bracket.

$$\begin{aligned}\Rightarrow 12a^2b + 15ab^2 \\[1em] \Rightarrow 3ab(4a + 5b).\end{aligned}$$

Hence, option 2 is the correct option.

Factorising,

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

Hence, option 2 is the correct option.

Factorising,

⇒ $49p^3q - 36pq$

⇒ $pq(49p^2 - 36)$

⇒ $pq((7p)^2 - (6)^2)$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $pq(7p + 6)(7p - 6)$

Hence, option 3 is the correct option.

Factorising,

$$\begin{aligned}\Rightarrow y(y - z) + 9(z - y) \\[1em] \Rightarrow y(y - z) - 9(y - z) \\[1em] \Rightarrow (y - z)(y - 9).\end{aligned}$$

Hence, option 3 is the correct option.

Factorising,

$$\begin{aligned}\Rightarrow (lm + l) + m + 1 \\[1em] \Rightarrow l(m + 1) + 1(m + 1) \\[1em] \Rightarrow (m + 1)(l + 1).\end{aligned}$$

Hence, option 4 is the correct option.

We want to find two integers whose sum is −4 and product is −12.

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

$$\begin{aligned}\Rightarrow z^2 - 4z - 12 \\[1em] \Rightarrow z^2 - 6z + 2z - 12 \\[1em] \Rightarrow z(z - 6) + 2(z - 6) \\[1em] \Rightarrow (z - 6)(z + 2).\end{aligned}$$

Hence, option 2 is the correct option.

Factorising,

⇒ $63a^2 - 112b^2$

⇒ $7(9a^2 - 16b^2)$

⇒ $7((3a)^2 - (4b)^2)$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $7(3a + 4b)(3a - 4b)$

Hence, option 3 is the correct option.

Factorising,

⇒ $p^4 - 81$

⇒ $(p^2)^2 - (9)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(p^2 + 9)(p^2 - 9)$

⇒ $(p^2 + 9)((p)^2 - (3)^2)$

⇒ $(p + 3)(p - 3)(p^2 + 9)$

Hence, option 3 is the correct option.

We want to find two integers whose sum is 7 and product is 3 × (−6) i.e. −18.

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

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

Hence, option 1 is the correct option.

Factorising,

⇒ $16x^2 + 40x + 25$

⇒ $(4x)^2 + 2 \times 4x \times 5 + (5)^2$

Using the identity,

[∵ $(x + y)^2 = x^2 + 2xy + y^2$]

⇒ $(4x + 5)^2$

⇒ $(4x + 5)(4x + 5)$

Hence, option 1 is the correct option.

Factorising,

⇒ $x^2 - 4xy + 4y^2$

⇒ $(x)^2 - 2 \times x \times 2y + (2y)^2$

Using the identity,

[∵ $(x - y)^2 = x^2 - 2xy + y^2$]

⇒ $(x - 2y)^2$

⇒ $(x - 2y)(x - 2y)$

Hence, option 2 is the correct option.

HCF of numerical coefficients = HCF of 6 and 12 = 6

The literal $y$ occurs in $6x^2y^3z$ but not in $12x^2z^3$, so it is not a common literal.

HCF of literal coefficients = HCF of $x^2y^3z$ and $x^2z^3$ = $x^2z$

HCF of the given monomials = 6 × $x^2z$ = $6x^2z$, which is not $6x^2yz$.

So, Statement I is false.

HCF of literal coefficients = product of each common literal raised to the lowest power, and not merely the product of each common literal.

So, Statement II is false.

Hence, option 4 is the correct option.

HCF of $3abc^2$ and $ab$ is $ab$

Divide each term by $ab$ and keep $ab$ outside the bracket.

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

So, the factors are $ab$ and $(3c^2 + 1)$, and not $(ab + 1)$.

So, Statement I is false.

Factorisation by grouping is possible only if the given polynomial contains an even number of terms.

Thus,

A polynomial with even number of terms can be factorised by grouping.

So, Statement II is true.

Hence, option 2 is the correct option.

Factorising,

⇒ $(x + 5)^2 - 1$

⇒ $(x + 5)^2 - (1)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $((x + 5) + 1)((x + 5) - 1)$

⇒ $(x + 6)(x + 4)$

So, Statement I is true.

$(a + b)(a - b) = a^2 - b^2$ is a standard algebraic identity.

So, Statement II is true.

Hence, option 3 is the correct option.

HCF of numerical coefficients = HCF of 14 and 28 = 14

HCF of literal coefficients = HCF of $pq$ and $p^2q^2$

= product of each common literal raised to the lowest power

= $pq$

HCF of the given monomials = 14 × $pq$ = $14pq$

Hence, the HCF of $14pq$ and $28p^2q^2$ is $\bm{14pq}$.

HCF of numerical coefficients = HCF of 8, 24 and 12 = 4

HCF of literal coefficients = HCF of $abc$, $ab^2$ and $a^2b$

= product of each common literal raised to the lowest power

= $ab$

HCF of the given monomials = 4 × $ab$ = $4ab$

Hence, the HCF of $8abc$, $24ab^2$ and $12a^2b$ is $\bm{4ab}$.

Factorising,

⇒ $10x^2 - 18x^3 + 14x^4$

HCF of $10x^2$, $18x^3$ and $14x^4$:

HCF of numerical coefficients = HCF of 10, 18 and 14 = 2

HCF of literal coefficients = HCF of $x^2$, $x^3$ and $x^4$

= product of each common literal raised to the lowest power

= $x^2$

∴ HCF of $10x^2$, $18x^3$ and $14x^4$ = 2 × $x^2$ = $2x^2$

Dividing each term by $2x^2$ and writing $2x^2$ outside the bracket,

⇒ $2x^2(5 - 9x + 7x^2)$

Hence, $\bm{10x^2 - 18x^3 + 14x^4 = 2x^2(5 - 9x + 7x^2)}$

Factorising,

⇒ $5x^2y + 10xyz + 15xy^2$

HCF of $5x^2y$, $10xyz$ and $15xy^2$:

HCF of numerical coefficients = HCF of 5, 10 and 15 = 5

HCF of literal coefficients = HCF of $x^2y$, $xyz$ and $xy^2$

= product of each common literal raised to the lowest power

= $xy$

∴ HCF of $5x^2y$, $10xyz$ and $15xy^2$ = 5 × $xy$ = $5xy$

Dividing each term by $5xy$ and writing $5xy$ outside the bracket,

⇒ $5xy(x + 2z + 3y)$

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

Factorising,

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

Hence, $\bm{p^2x^2 + c^2x^2 - ac^2 - ap^2 = (p^2 + c^2)(x^2 - a)}$

Factorising,

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

Hence, $\bm{15(x + y)^2 - 5x - 5y = 5(x + y)(3x + 3y - 1)}$

Factorising,

⇒ $(ax + by)^2 + (ay - bx)^2$

Using the identity,

[∵ $(x + y)^2 = x^2 + 2xy + y^2$]

And

[∵ $(x - y)^2 = x^2 - 2xy + y^2$]

⇒ $(a^2x^2 + 2abxy + b^2y^2) + (a^2y^2 - 2abxy + b^2x^2)$

⇒ $a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 + 2abxy - 2abxy$

⇒ $a^2(x^2 + y^2) + b^2(x^2 + y^2)$

⇒ $(x^2 + y^2)(a^2 + b^2)$

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

Factorising,

$$\begin{aligned}\Rightarrow ax + by + cx + bx + cy + ay \\[1em] \Rightarrow (ax + bx + cx) + (ay + by + cy) \\[1em] \Rightarrow x(a + b + c) + y(a + b + c) \\[1em] \Rightarrow (x + y)(a + b + c).\end{aligned}$$

Hence, $\bm{ax + by + cx + bx + cy + ay = (x + y)(a + b + c)}$

Factorising,

⇒ $49x^2 - 70xy + 25y^2$

⇒ $(7x)^2 - 2 \times 7x \times 5y + (5y)^2$

Using the identity,

[∵ $(x - y)^2 = x^2 - 2xy + y^2$]

⇒ $(7x - 5y)^2$

⇒ $(7x - 5y)(7x - 5y)$

Hence, $\bm{49x^2 - 70xy + 25y^2 = (7x - 5y)(7x - 5y) = (7x - 5y)^2}$

Factorising,

⇒ $4a^2 + 12ab + 9b^2$

⇒ $(2a)^2 + 2 \times 2a \times 3b + (3b)^2$

Using the identity,

[∵ $(x + y)^2 = x^2 + 2xy + y^2$]

⇒ $(2a + 3b)^2$

⇒ $(2a + 3b)(2a + 3b)$

Hence, $\bm{4a^2 + 12ab + 9b^2 = (2a + 3b)(2a + 3b) = (2a + 3b)^2}$

Factorising,

⇒ $49p^2 - 36q^2$

⇒ $(7p)^2 - (6q)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(7p + 6q)(7p - 6q)$

Hence, $\bm{49p^2 - 36q^2 = (7p + 6q)(7p - 6q)}$

Factorising,

⇒ $100x^3 - 25xy^2$

⇒ $25x(4x^2 - y^2)$

⇒ $25x((2x)^2 - (y)^2)$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $25x(2x + y)(2x - y)$

Hence, $\bm{100x^3 - 25xy^2 = 25x(2x + y)(2x - y)}$

Factorising,

⇒ $x^2 - 2xy + y^2 - z^2$

⇒ $(x^2 - 2xy + y^2) - z^2$

Using the identity,

[∵ $(x - y)^2 = x^2 - 2xy + y^2$]

⇒ $(x - y)^2 - (z)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $((x - y) + z)((x - y) - z)$

⇒ $(x - y + z)(x - y - z)$

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

Factorising,

⇒ $x^8 - y^8$

⇒ $(x^4)^2 - (y^4)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(x^4 + y^4)(x^4 - y^4)$

⇒ $(x^4 + y^4)((x^2)^2 - (y^2)^2)$

⇒ $(x^4 + y^4)(x^2 + y^2)(x^2 - y^2)$

⇒ $(x^4 + y^4)(x^2 + y^2)(x + y)(x - y)$

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

HCF of $12x^3$, $14x^2$ and $10x$ is $2x$

Divide each term by $2x$ and keep $2x$ outside the bracket.

⇒ $12x^3 - 14x^2 - 10x$

⇒ $2x(6x^2 - 7x - 5)$

We want to find two integers whose sum is −7 and product is 6 × (−5) i.e. −30.

By trial, we see that (−10) + 3 = −7 and (−10) × 3 = −30

⇒ $2x(6x^2 - 10x + 3x - 5)$

⇒ $2x(2x(3x - 5) + 1(3x - 5))$

⇒ $2x(3x - 5)(2x + 1)$

Hence, $\bm{12x^3 - 14x^2 - 10x = 2x(3x - 5)(2x + 1)}$

We want to find two integers whose sum is −10 and product is 21.

By trial, we see that (−7) + (−3) = −10 and (−7) × (−3) = 21

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

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

We want to find two integers whose sum is −1 and product is 2 × (−6) i.e. −12.

By trial, we see that 3 + (−4) = −1 and 3 × (−4) = −12

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

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

We want to find two integers whose sum is −5 and product is 6 × (−6) i.e. −36.

By trial, we see that (−9) + 4 = −5 and (−9) × 4 = −36

$$\begin{aligned}\Rightarrow 6x^2 - 5xy - 6y^2 \\[1em] \Rightarrow 6x^2 - 9xy + 4xy - 6y^2 \\[1em] \Rightarrow 3x(2x - 3y) + 2y(2x - 3y) \\[1em] \Rightarrow (2x - 3y)(3x + 2y).\end{aligned}$$

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

We want to find two integers whose sum is 2 and product is −99.

By trial, we see that 11 + (−9) = 2 and 11 × (−9) = −99

$$\begin{aligned}\Rightarrow x^2 + 2xy - 99y^2 \\[1em] \Rightarrow x^2 + 11xy - 9xy - 99y^2 \\[1em] \Rightarrow x(x + 11y) - 9y(x + 11y) \\[1em] \Rightarrow (x + 11y)(x - 9y).\end{aligned}$$

Hence, $\bm{x^2 + 2xy - 99y^2 = (x + 11y)(x - 9y)}$

Factorising $y^2 - 16$, we get

⇒ $y^2 - 16$

⇒ $(y)^2 - (4)^2$

⇒ $(y + 4)(y - 4)$

Factorising $y^2 - y - 12$, we want two integers whose sum is −1 and product is −12.

By trial, we see that (−4) + 3 = −1 and (−4) × 3 = −12

⇒ $y^2 - y - 12$

⇒ $y^2 - 4y + 3y - 12$

⇒ $y(y - 4) + 3(y - 4)$

⇒ $(y - 4)(y + 3)$

Now,

$$\begin{aligned}\Rightarrow \frac{15(y + 3)(y^2 - 16)}{5(y^2 - y - 12)}\\[1em] \Rightarrow \frac{15(y + 3)(y + 4)(y - 4)}{5(y - 4)(y + 3)}\\[1em] \Rightarrow \frac{15\cancel{(y + 3)}(y + 4)\cancel{(y - 4)}}{5\cancel{(y - 4)}\cancel{(y + 3)}}\\[1em] \Rightarrow \frac{15(y + 4)}{5}\\[1em] \Rightarrow 3(y + 4)\end{aligned}$$

Hence, $\bm{15(y + 3)(y^2 - 16) \div 5(y^2 - y - 12) = 3(y + 4)}$

Factorising the numerator $3x^3 - 6x^2 - 24x$, we get

⇒ $3x^3 - 6x^2 - 24x$

⇒ $3x(x^2 - 2x - 8)$

We want to find two integers whose sum is −2 and product is −8.

By trial, we see that (−4) + 2 = −2 and (−4) × 2 = −8

⇒ $3x(x^2 - 2x - 8)$

⇒ $3x(x^2 - 4x + 2x - 8)$

⇒ $3x(x(x - 4) + 2(x - 4))$

⇒ $3x(x - 4)(x + 2)$

Now,

$$\begin{aligned}\Rightarrow \frac{3x^3 - 6x^2 - 24x}{(x - 4)(x + 2)}\\[1em] \Rightarrow \frac{3x(x - 4)(x + 2)}{(x - 4)(x + 2)}\\[1em] \Rightarrow \frac{3x\cancel{(x - 4)}\cancel{(x + 2)}}{\cancel{(x - 4)}\cancel{(x + 2)}}\\[1em] \Rightarrow 3x\end{aligned}$$

Hence, $\bm{(3x^3 - 6x^2 - 24x) \div (x - 4)(x + 2) = 3x}$

Factorising the numerator $x^4 - 81$, we get

⇒ $x^4 - 81$

⇒ $(x^2)^2 - (9)^2$

⇒ $(x^2 + 9)(x^2 - 9)$

⇒ $(x^2 + 9)(x + 3)(x - 3)$

Factorising the denominator $x^3 + 3x^2 + 9x + 27$, we get

⇒ $x^3 + 3x^2 + 9x + 27$

⇒ $x^2(x + 3) + 9(x + 3)$

⇒ $(x + 3)(x^2 + 9)$

Now,

$$\begin{aligned}\Rightarrow \frac{x^4 - 81}{x^3 + 3x^2 + 9x + 27}\\[1em] \Rightarrow \frac{(x^2 + 9)(x + 3)(x - 3)}{(x + 3)(x^2 + 9)}\\[1em] \Rightarrow \frac{\cancel{(x^2 + 9)}\cancel{(x + 3)}(x - 3)}{\cancel{(x + 3)}\cancel{(x^2 + 9)}}\\[1em] \Rightarrow x - 3\end{aligned}$$

Hence, $\bm{(x^4 - 81) \div (x^3 + 3x^2 + 9x + 27) = x - 3}$

Here the sum of the two parts is $a + \frac{1}{a}$ and their product is $a \times \frac{1}{a}$ = 1, so we split the middle term into $ax$ and $\frac{1}{a}x$.

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

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

We want to find two integers whose sum is −97 and product is 36 × 36 i.e. 1296.

By trial, we see that (−81) + (−16) = −97 and (−81) × (−16) = 1296

⇒ $36a^4 - 97a^2b^2 + 36b^4$

⇒ $36a^4 - 81a^2b^2 - 16a^2b^2 + 36b^4$

⇒ $9a^2(4a^2 - 9b^2) - 4b^2(4a^2 - 9b^2)$

⇒ $(4a^2 - 9b^2)(9a^2 - 4b^2)$

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

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $(2a + 3b)(2a - 3b)(3a + 2b)(3a - 2b)$

Hence, $\bm{36a^4 - 97a^2b^2 + 36b^4 = (3a + 2b)(3a - 2b)(2a - 3b)(2a + 3b)}$

Here the product of the two parts is $2 \times (-3)$ = $-6$ and their sum is $-\sqrt{3}$.

By trial, we see that $−2\sqrt{3} + \sqrt{3} = -\sqrt{3}$ and $−2\sqrt{3} \times \sqrt{3} = −6$

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

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

Factorising,

⇒ $y(y^2 - 2y) + 2(2y - y^2) - 2 + y$

⇒ $y(y^2 - 2y) - 2(y^2 - 2y) + (y - 2)$

⇒ $(y^2 - 2y)(y - 2) + 1(y - 2)$

⇒ $(y - 2)(y^2 - 2y + 1)$

Using the identity,

[∵ $(x - y)^2 = x^2 - 2xy + y^2$]

⇒ $(y - 2)(y - 1)^2$

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