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