Factorisation — Exercise 11.4

Factorise the following polynomial:

$x^2 + 3x + 2$

Answer

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

$\Rightarrow x^2 + 3x + 2 \\[1em] \Rightarrow x^2 + (1 + 2)x + 2 \\[1em] \Rightarrow x^2 + x + 2x + 2 \\[1em] \Rightarrow x(x + 1) + 2(x + 1) \\[1em] \Rightarrow (x + 1)(x + 2).$

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

Factorise the following polynomial:

$z^2 + 10z + 24$

Answer

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

$\Rightarrow z^2 + 10z + 24 \\[1em] \Rightarrow z^2 + (4 + 6)z + 24 \\[1em] \Rightarrow z^2 + 4z + 6z + 24 \\[1em] \Rightarrow z(z + 4) + 6(z + 4) \\[1em] \Rightarrow (z + 4)(z + 6).$

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

Factorise the following polynomial:

$y^2 - 7y + 12$

Answer

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

$\Rightarrow y^2 - 7y + 12 \\[1em] \Rightarrow y^2 - (3 + 4)y + 12 \\[1em] \Rightarrow y^2 - 3y - 4y + 12 \\[1em] \Rightarrow y(y - 3) - 4(y - 3) \\[1em] \Rightarrow (y - 3)(y - 4).$

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

Factorise the following polynomial:

$m^2 - 23m + 42$

Answer

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

$\Rightarrow m^2 - 23m + 42 \\[1em] \Rightarrow m^2 - (2 + 21)m + 42 \\[1em] \Rightarrow m^2 - 2m - 21m + 42 \\[1em] \Rightarrow m(m - 2) - 21(m - 2) \\[1em] \Rightarrow (m - 2)(m - 21).$

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

Factorise the following polynomial:

$y^2 - 5y - 24$

Answer

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

$\Rightarrow y^2 - 5y - 24 \\[1em] \Rightarrow y^2 + (3 - 8)y - 24 \\[1em] \Rightarrow y^2 + 3y - 8y - 24 \\[1em] \Rightarrow y(y + 3) - 8(y + 3) \\[1em] \Rightarrow (y + 3)(y - 8).$

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

Factorise the following polynomial:

$t^2 + 23t - 108$

Answer

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

$\Rightarrow t^2 + 23t - 108 \\[1em] \Rightarrow t^2 + (27 - 4)t - 108 \\[1em] \Rightarrow t^2 + 27t - 4t - 108 \\[1em] \Rightarrow t(t + 27) - 4(t + 27) \\[1em] \Rightarrow (t + 27)(t - 4).$

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

Factorise the following polynomial:

$3x^2 + 14x + 8$

Answer

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

$\Rightarrow 3x^2 + 14x + 8 \\[1em] \Rightarrow 3x^2 + 12x + 2x + 8 \\[1em] \Rightarrow 3x(x + 4) + 2(x + 4) \\[1em] \Rightarrow (x + 4)(3x + 2).$

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

Factorise the following polynomial:

$3y^2 + 10y + 8$

Answer

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

$\Rightarrow 3y^2 + 10y + 8 \\[1em] \Rightarrow 3y^2 + 6y + 4y + 8 \\[1em] \Rightarrow 3y(y + 2) + 4(y + 2) \\[1em] \Rightarrow (y + 2)(3y + 4).$

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

Factorise the following polynomial:

$14x^2 - 23x + 8$

Answer

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

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

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

Factorise the following polynomial:

$12x^2 - x - 35$

Answer

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

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

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

Factorise the following polynomial:

$6x^2 + 11x - 10$

Answer

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

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

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

Factorise the following polynomial:

$5 - 4x - 12x^2$

Answer

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

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

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

Factorise the following polynomial:

$1 - 18y - 63y^2$

Answer

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

$\Rightarrow 1 - 18y - 63y^2 \\[1em] \Rightarrow 1 - 21y + 3y - 63y^2 \\[1em] \Rightarrow 1(1 - 21y) + 3y(1 - 21y) \\[1em] \Rightarrow (1 - 21y)(1 + 3y).$

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

Factorise the following polynomial:

$3x^2 - 5xy - 12y^2$

Answer

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

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

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

Factorise the following polynomial:

$x^2 - 3xy - 40y^2$

Answer

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

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

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

Factorise the following polynomial:

$10p^2q^2 - 21pq + 9$

Answer

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

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

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

Factorise the following polynomial:

$2a^2b^2 + ab - 45$

Answer

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

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

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

Factorise the following polynomial:

$x(12x + 7) - 10$

Answer

Solving,

$\Rightarrow x(12x + 7) - 10 \\[1em] \Rightarrow 12x^2 + 7x - 10$

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

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

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

Factorise the following polynomial:

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

Answer

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

$\Rightarrow m^2 - 11m - 42 \\[1em] \Rightarrow m^2 + 3m - 14m - 42 \\[1em] \Rightarrow m(m + 3) - 14(m + 3) \\[1em] \Rightarrow (m + 3)(m - 14).$

Substituting $m = a + b$, we get

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

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

Factorise the following polynomial:

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

Answer

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

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

Substituting $m = p + q$, we get

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

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

Factorise the following polynomial:

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

Answer

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

$\Rightarrow m^2 - 6m + 5 \\[1em] \Rightarrow m^2 - m - 5m + 5 \\[1em] \Rightarrow m(m - 1) - 5(m - 1) \\[1em] \Rightarrow (m - 1)(m - 5).$

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

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

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

Factorise the following polynomial:

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

Answer

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

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

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

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

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