Prove the following

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]

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

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