Factorisation — Exercise 11.5

Work out the following division:

$(35x + 28) \div (5x + 4)$

Answer

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

⇒ $(35x + 28)$

⇒ $7(5x + 4)$

Solving,

$\Rightarrow \dfrac{35x + 28}{5x + 4}\\[1em] \Rightarrow \dfrac{7(5x + 4)}{(5x + 4)}\\[1em] \Rightarrow \dfrac{7\cancel{(5x + 4)}}{\cancel{(5x + 4)}}\\[1em] \Rightarrow 7$

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

Work out the following division:

$7p^2q^2(9r - 27) \div 63pq(r - 3)$

Answer

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

Solving,

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

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

Divide as directed:

$6(2x + 7)(5x - 3) \div 3(5x - 3)$

Answer

Solving,

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

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

Divide as directed:

$33pq(p + 3)(2q - 5) \div 11p(2q - 5)$

Answer

Solving,

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

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

Factorise the expression and divide them as directed:

$(7x^3 - 63x) \div 7(x - 3)$

Answer

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,

$\Rightarrow \dfrac{7x^3 - 63x}{7(x - 3)}\\[1em] \Rightarrow \dfrac{7x(x + 3)(x - 3)}{7(x - 3)}\\[1em] \Rightarrow \dfrac{\cancel{7}x(x + 3)\cancel{(x - 3)}}{\cancel{7}\cancel{(x - 3)}}\\[1em] \Rightarrow x(x + 3)$

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

Factorise the expression and divide them as directed:

$(3p^2 + 17p + 10) \div (p + 5)$

Answer

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,

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

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

Factorise the expression and divide them as directed:

$10xy (14y^2 + 43y - 21) \div 5x(7y - 3)$

Answer

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,

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

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

Factorise the expression and divide them as directed:

$12pqr(6p^2 - 13pq + 6q^2) \div 6pq(2p - 3q)$

Answer

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,

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

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