Question

Question 24
Factorise
(i) $2x^3–3x^2–17x+30$

(ii) $x^3–6x^2+11x–6$

(iii) $x^3+x^2-4x-4$

(iv) $3x^3–x^2-3x+1$

Answer 1

(i) Let $p\left(x\right)=2x^3–3x^2–17x+30$
Constant term of p(x) = 30
$\therefore$ Factors of 30 are $\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30$
By trial, we find that p(2) = 0, so (x – 2) is a factor of p(x).
$[\because 2(2{)}^3-3(2{)}^2-17(2)+30=16-12-34+30=0]$


Now, we see that $2x^3–3x^2–17x+30$

$=(x-2)(2x^2+x-15)$
$2x^2+x-15=2x(x+3)–5(x+3)$ [By splitting the middle term]
$=(x+3)(2x–5)$
$\therefore 2x^3–3x^2–17x+30=(x-2)(x+3)(2x-5)$

(ii) Let $p\left(x\right)=x^3–6x^2+11x–6$
Constant term of p(x) = - 6
Factors of -6 are $\pm 1,\pm 2,\pm 3,\pm 6$.
By trial, we find that p(1) = 0. So, ( x – 1 ) is a factor of p(x).
$[\because (1{)}^3-6(1{)}^2+11(1)-6=1-6+11-6=0]$
Now, $x^3–6x^2+11x–6$
$=x^3–x^2–5x^2+5x+6x–6$
$=(x-1)(x^2–5x+6)$ [Taking (x – 1) common factor]
Now, $(x^2–5x+6)=x^2-3x–2x+6$ [By splitting the middle term]
$=x(x–3)–2(x–3)$
$=(x–3)(x–2)$
$\therefore x^3–6x^2+11x–6=(x–1)(x–2)(x–3)$

(iii) Let $p\left(x\right)=x^3+x^2-4x-4$
Constant term of p(x) = - 4
Factors of -4 are $\pm 1,\pm 2,\pm 4$.
By trial, we find that p(-1) = 0. So, (x + 1) is a factor of p(x).
Now,
$x^3+x^2-4x-4$
$=x^2(x+1)–4(x+1)$
$=(x+1)(x^2–4)$ [Taking (x + 1) common factor]
Now, $x^2–4=x^2–2^2$
$=(x+2)(x–2)$ [Using identity, $a^2–b^2=(a–b)(a+b)]$
$\therefore x^3+x^2–4x–4=(x+1)(x–2)(x+2)$

(iv) Let $p\left(x\right)=3x^3–x^2-3x+1$
Constant term of p(x) = 1
Factor of 1 are 1.
By trial, we find that p(1) = 0 , so (x – 1) is a factor of p(x).
Now,
$3x^3–x^2-3x+1$
$=3x^3-3x^2+2x^2-2x–x+1$
$=3x^2(x–1)+2x(x–1)–1(x–1)$
$=(x–1)(3x^2+2x–1)$
Now, $(3x^2+2x–1)=3x^2+3x-x-1$ [By spitting the middle term]
$=3x(x+1)-1(x+1)=(x+1)(3x-1)$
$\therefore 3x^3–x^2-3x+1=(x-1)(x+1)(3x-1)$