Factorisation — Exercise 11.2

Factorise the following polynomial:

$x^2 + xy - x - y$

Answer

Factorising,

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

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

Factorise the following polynomial:

$y^2 - yz - 5y + 5z$

Answer

Factorising,

$\Rightarrow y^2 - yz - 5y + 5z \\[1em] \Rightarrow (y^2 - yz) - (5y - 5z) \\[1em] \Rightarrow y(y - z) - 5(y - z) \\[1em] \Rightarrow (y - z)(y - 5).$

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

Factorise the following polynomial:

$5xy + 7y - 5y^2 - 7x$

Answer

Factorising,

$\Rightarrow 5xy + 7y - 5y^2 - 7x \\[1em] \Rightarrow 5xy - 5y^2 + 7y - 7x \\[1em] \Rightarrow (5xy - 5y^2) + (7y - 7x) \\[1em] \Rightarrow 5y(x - y) - 7(x - y) \\[1em] \Rightarrow (x - y)(5y - 7).$

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

Factorise the following polynomial:

$5p^2 - 8pq - 10p + 16q$

Answer

Factorising,

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

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

Factorise the following polynomial:

$a^2b - ab^2 + 3a - 3b$

Answer

Factorising,

$\Rightarrow a^2b - ab^2 + 3a - 3b \\[1em] \Rightarrow (a^2b - ab^2) + (3a - 3b) \\[1em] \Rightarrow ab(a - b) + 3(a - b) \\[1em] \Rightarrow (a - b)(ab + 3).$

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

Factorise the following polynomial:

$x^3 - 3x^2 + x - 3$

Answer

Factorising,

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

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

Factorise the following polynomial:

$6xy^2 - 3xy - 10y + 5$

Answer

Factorising,

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

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

Factorise the following polynomial:

$3ax - 6ay - 8by + 4bx$

Answer

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

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

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

Factorise the following polynomial:

$x^2 + xy(1 + y) + y^3$

Answer

Factorising,

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

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

Factorise the following polynomial:

$y^2 - xy(1 - x) - x^3$

Answer

Factorising,

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

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

Factorise the following polynomial:

$ab^2 + (a - 1)b - 1$

Answer

Factorising,

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

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

Factorise the following polynomial:

$2a - 4b - xa + 2bx$

Answer

Factorising,

$\Rightarrow 2a - 4b - xa + 2bx \\[1em] \Rightarrow (2a - 4b) - (xa - 2bx) \\[1em] \Rightarrow 2(a - 2b) - x(a - 2b) \\[1em] \Rightarrow (a - 2b)(2 - x).$

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

Factorise the following polynomial:

$5ph - 10qk + 2rph - 4qrk$

Answer

Factorising,

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

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

Factorise the following polynomial:

$x^2 - x(a + 2b) + 2ab$

Answer

Factorising,

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

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

Factorise the following polynomial:

$ab(x^2 + y^2) - xy(a^2 + b^2)$

Answer

Factorising,

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

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

Factorise the following polynomial:

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

Answer

Using the identity,

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

Factorising,

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

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

Factorise the following polynomial:

$a^3 + ab(1 - 2a) - 2b^2$

Answer

Factorising,

$\Rightarrow a^3 + ab(1 - 2a) - 2b^2 \\[1em] \Rightarrow a^3 + ab - 2a^2b - 2b^2 \\[1em] \Rightarrow (a^3 - 2a^2b) + (ab - 2b^2) \\[1em] \Rightarrow a^2(a - 2b) + b(a - 2b) \\[1em] \Rightarrow (a - 2b)(a^2 + b).$

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

Factorise the following polynomial:

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

Answer

Factorising,

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

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

Factorise the following polynomial:

$a^2b + ab^2 - abc - b^2c + axy + bxy$

Answer

Factorising,

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

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

Factorise the following polynomial:

$ax^2 - bx^2 + ay^2 - by^2 + az^2 - bz^2$

Answer

Factorising,

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

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

Factorise the following polynomial:

$x - 1 - (x - 1)^2 + ax - a$

Answer

Factorising,

$\Rightarrow x - 1 - (x - 1)^2 + ax - a \\[1em] \Rightarrow (x - 1) - (x - 1)^2 + (ax - a) \\[1em] \Rightarrow 1(x - 1) - (x - 1)^2 + a(x - 1) \\[1em] \Rightarrow (x - 1)[1 - (x - 1) + a] \\[1em] \Rightarrow (x - 1)(1 - x + 1 + a) \\[1em] \Rightarrow (x - 1)(2 - x + a).$

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

Factorise the following polynomial:

$ax + a^2x + aby + by - (ax + by)^2$

Answer

Factorising,

$\Rightarrow ax + a^2x + aby + by - (ax + by)^2 \\[1em] \Rightarrow (ax + a^2x) + (aby + by) - (ax + by)^2 \\[1em] \Rightarrow ax(1 + a) + by(a + 1) - (ax + by)^2 \\[1em] \Rightarrow (a + 1)(ax + by) - (ax + by)^2 \\[1em] \Rightarrow (ax + by)[(a + 1) - (ax + by)] \\[1em] \Rightarrow (ax + by)(a + 1 - ax - by) \\[1em] \Rightarrow (ax + by)(1 + a - ax - by).$

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