# Prove the following

Question:

Factorise

(i) 2x3 -3x2 -17x + 30

(ii) x3 -6x2 +11 x-6

(iii) x3 + x2 – 4x – 4

(iv) 3x3 – x2 – 3x +1

Thinking Process

(i) Firstly, find the prime factors of constant term in given polynomial.

(ii) Substitute these factors in place of x in given polynomial and find the minimum factor which satisfies given polynomial and write it in the form of a

linear polynomial.

(iii) Now, adjust the given polynomial in such a way that it becomes the product of two factors, one of them is a linear polynomial and other is a

quadratic polynomial.

(iv) Further, determine the factor of quadratic polynomial by splitting the middle term.

Solution:

(i) Let $p(x)=2 x^{3}-3 x^{2}-17 x+30$

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 $p(x)$.

$\left[\because 2(2)^{3}-3(2)^{2}-17(2)+30=16-12-34+30=0\right]$

Now, we see that $2 x^{3}-3 x^{2}-17 x+30$

$=2 x^{3}-4 x^{2}+x^{2}-2 x-15 x+30$

$=2 x^{2}(x-2)+x(x-2)-15(x-2)$

$=(x-2)\left(2 x^{2}+x-15\right) \quad$ [taking $(x-2)$ common factor $]$

Now, $\left(2 x^{2}+x-15\right)$ can be factorised either by splitting the middle term or by using the factor theorem.

Now, $\quad\left(2 x^{2}+x-15\right)=2 x^{2}+6 x-5 x-15$  [by splitting the middle term]

$=2 x(x+3)-5(x+3)$

$=(x+3)(2 x-5)$

$\therefore$$2 x^{3}-3 x^{2}-17 x+30=(x-2)(x+3)(2 x-5) (ii) Let p(x)=x^{3}-6 x^{2}+11 x-6 Constant term of p(x)=-6 Factors of -6 are \pm 1, \pm 2, \pm 3, \pm 6. By trial, we find that p(1)=0, so (x-1) is a factor of p(x). \left[\because(1)^{3}-6(1)^{2}+11(1)-6=1-6+11-6=0\right] Now, we see that x^{3}-6 x^{2}+11 x-6 =x^{3}-x^{2}-5 x^{2}+5 x+6 x-6 \left.\left.=x^{2}(x-1)-5 x x-1\right)+6 x-1\right) =(x-1)\left(x^{2}-5 x+6\right) [taking (x-1) common factor] Now, \quad\left(x^{2}-5 x+6\right)=x^{2}-3 x-2 x+6 [by splitting the middle term] =x(x-3)-2(x-3) =(x-3)(x-2) \therefore \quad x^{3}-6 x^{2}+11 x-6=(x-1)(x-2)(x-3) (iii) Let p(x)=x^{3}+x^{2}-4 x-4 Constant term of p(x)=-4 Factors of -4 are \pm 1, \pm 2, \pm 4. By trial, we find that p(-1)=0, so (x+1) is a factor of p(x). Now, we see that x^{3}+x^{2}-4 x-4 =x^{2}(x+1)-4(x+1) =(x+1)\left(x^{2}-4\right)$$\quad[$ taking $(x+1)$ common factor $]$

Now, $x^{2}-4=x^{2}-2^{2}$

$=(x+2)(x-2) \quad$ [using identity, $\left.a^{2}-b^{2}=(a-b)(a+b)\right]$

$\therefore \quad x^{3}+x^{2}-4 x-4=(x+1)(x-2)(x+2)$

(lv) Let $p(x)=3 x^{3}-x^{2}-3 x+1$

Constant term of $p(x)=1$

Factor of 1 are $\pm 1$.

By trial, we find that $p(1)=0$, so $(x-1)$ is a factor of $p(x)$.

Now, we see that $3 x^{3}-x^{2}-3 x+1$

$=3 x^{3}-3 x^{2}+2 x^{2}-2 x-x+1$

$=3 x^{2}(x-1)+2 x(x-1)-1(x-1)$

$=(x-1)\left(3 x^{2}+2 x-1\right)$

Now, $\quad\left(3 x^{2}+2 x-1\right)=3 x^{2}+3 x-x-1$   [by splitting middle term]

$=3 x(x+1)-1(x+1)=(x+1)(3 x-1)$

$\therefore \quad 3 x^{3}-x^{2}-3 x+1=(x-1)(x+1)(3 x-1)$

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