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# Question 24 Factorise (i) 2x3–3x2–17x+30 (ii) x3–6x2+11x–6 (iii) x3+x2−4x−4 (iv) 3x3–x2−3x+1

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Solution

## (i) Let p(x)=2x3–3x2–17x+30 Constant term of p(x) = 30 ∴ Factors of 30 are ±1,±2,±3,±5,±6,±10,±15,±30 By trial, we find that p(2) = 0, so (x – 2) is a factor of p(x). [∵2(2)3−3(2)2−17(2)+30=16−12−34+30=0] Now, we see that 2x3–3x2–17x+30 =(x−2)(2x2+x−15) 2x2+x−15=2x(x+3)–5(x+3) [By splitting the middle term] =(x+3)(2x–5) ∴ 2x3–3x2–17x+30=(x−2)(x+3)(2x−5) (ii) Let p(x)=x3–6x2+11x–6 Constant term of p(x) = - 6 Factors of -6 are ±1,±2,±3,±6. By trial, we find that p(1) = 0. So, ( x – 1 ) is a factor of p(x). [∵ (1)3−6(1)2+11(1)−6=1−6+11−6=0] Now, x3–6x2+11x–6 =x3–x2–5x2+5x+6x–6 =(x−1)(x2–5x+6) [Taking (x – 1) common factor] Now, (x2–5x+6)=x2−3x–2x+6 [By splitting the middle term] =x(x–3)–2(x–3) =(x–3)(x–2) ∴ x3–6x2+11x–6=(x–1)(x–2)(x–3) (iii) Let p(x)=x3+x2−4x−4 Constant term of p(x) = - 4 Factors of -4 are ±1,±2,±4. By trial, we find that p(-1) = 0. So, (x + 1) is a factor of p(x). Now, x3+x2−4x−4 =x2(x+1)–4(x+1) =(x+1)(x2–4) [Taking (x + 1) common factor] Now, x2–4=x2–22 =(x+2)(x–2) [Using identity, a2–b2=(a–b)(a+b)] ∴ x3+x2–4x–4=(x+1)(x–2)(x+2) (iv) Let p(x)=3x3–x2−3x+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, 3x3–x2−3x+1 =3x3−3x2+2x2−2x–x+1 =3x2(x–1)+2x(x–1)–1(x–1) =(x–1)(3x2+2x–1) Now, (3x2+2x–1)=3x2+3x−x−1 [By spitting the middle term] =3x(x+1)−1(x+1)=(x+1)(3x−1) ∴ 3x3–x2−3x+1=(x−1)(x+1)(3x−1)

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