Question

Factorise $f\left(x\right)=x^3+2x^2-5x-6$ using synthetic division and find zeros of it.

• A. $-2,-3,-1$
• B. $2,3,1$
• C. $-2,3,1$
• D. $2,-3,-1$

Answer 1

The correct option is D $2,-3,-1$

Let $f\left(x\right)=x^3+2x^2-x-6$

We find the first zero by hit and trial method

So Put $x=-1$

$f(-1)=-1+2+5-6=0$

So $(x+1)$ is a factor of $f\left(x\right)$

Dividing $f\left(x\right)$ by $(x+1)$ we get

$p\left(x\right)=x^2+x-6$

Factorising $p\left(x\right)$

$p\left(x\right)=x^2+3x-2x-6p\left(x\right)=x(x+3)-2(x+3)p\left(x\right)=(x-2)(x+3)$

$f\left(x\right)=(x+1)p\left(x\right)$

$f\left(x\right)=(x+1)(x-2)(x+3)$

So the zeroes are $x=-1,2,-3$

Option $D$ is correct.