Factorisation — Check Your Progress

Find the HCF of the given polynomial:

$14pq$, $28p^2q^2$

Answer

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 $\bm{14pq}$.

Find the HCF of the given polynomial:

$8abc$, $24ab^2$, $12a^2b$

Answer

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 $\bm{4ab}$.

Factorise the following:

$10x^2 - 18x^3 + 14x^4$

Answer

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

Factorise the following:

$5x^2y + 10xyz + 15xy^2$

Answer

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

Factorise the following:

$p^2x^2 + c^2x^2 - ac^2 - ap^2$

Answer

Factorising,

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

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

Factorise the following:

$15(x + y)^2 - 5x - 5y$

Answer

Factorising,

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

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

Factorise the following:

$(ax + by)^2 + (ay - bx)^2$

Answer

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

Factorise the following:

$ax + by + cx + bx + cy + ay$

Answer

Factorising,

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

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

Factorise the following:

$49x^2 - 70xy + 25y^2$

Answer

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

Factorise the following:

$4a^2 + 12ab + 9b^2$

Answer

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

Factorise the following:

$49p^2 - 36q^2$

Answer

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

Factorise the following:

$100x^3 - 25xy^2$

Answer

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

Factorise the following:

$x^2 - 2xy + y^2 - z^2$

Answer

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

Factorise the following:

$x^8 - y^8$

Answer

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

Factorise the following:

$12x^3 - 14x^2 - 10x$

Answer

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

Factorise the following:

$p^2 - 10p + 21$

Answer

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

$\Rightarrow p^2 - 10p + 21 \\[1em] \Rightarrow p^2 - 7p - 3p + 21 \\[1em] \Rightarrow p(p - 7) - 3(p - 7) \\[1em] \Rightarrow (p - 7)(p - 3).$

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

Factorise the following:

$2x^2 - x - 6$

Answer

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

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

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

Factorise the following:

$6x^2 - 5xy - 6y^2$

Answer

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

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

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

Factorise the following:

$x^2 + 2xy - 99y^2$

Answer

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

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

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

Divide as directed:

$15(y + 3) (y^2 - 16) \div 5(y^2 - y - 12)$

Answer

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,

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

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

Divide as directed:

$(3x^3 - 6x^2 - 24x) \div (x - 4) (x + 2)$

Answer

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,

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

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

Divide as directed:

$(x^4 - 81) \div (x^3 + 3x^2 + 9x + 27)$

Answer

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,

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

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

Factorise the following:

$x^2 + \left(a + \dfrac{1}{a}\right)x + 1$

Answer

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

$\Rightarrow x^2 + \Big(a + \dfrac{1}{a}\Big)x + 1\\[1em] \Rightarrow x^2 + ax + \dfrac{1}{a}x + 1\\[1em] \Rightarrow x(x + a) + \dfrac{1}{a}\Big(x + a\Big)\\[1em] \Rightarrow (x + a)\Big(x + \dfrac{1}{a}\Big)$

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

Factorise the following:

$36a^4 - 97a^2b^2 + 36b^4$

Answer

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

Factorise the following:

$2x^2 - \sqrt{3}x - 3$

Answer

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 $−2\sqrt{3} \times \sqrt{3} = −6$

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

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

Factorise the following:

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

Answer

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