Question

Factorise the following polynomial. $x^3-2x^2-5x+6$

• A. $(x-1)(x+2)(x-3)$
• B. $(x-1)(x-2)(x-3)$
• C. $(x-1)(x+2)(x+3)$
• D. None of these

Answer 1

The correct option is A $(x-1)(x+2)(x-3)$

Let the given polynomial be $p\left(x\right)=x^3-2x^2-5x+6$.

We will now substitute various values of $x$ until we get $p\left(x\right)=0$ as follows:

$Forx=0p\left(0\right)=(0{)}^3-2(0{)}^2-(5\times 0)+6=0-0-0+6=6\ne 0\therefore p\left(0\right)\ne 0$

$Forx=1p\left(1\right)=(1{)}^3-2(1{)}^2-(5\times 1)+6=1-2-5+6=7-7=0\therefore p\left(1\right)=0$

Thus, $(x-1)$ is a factor of $p\left(x\right)$.

Now,

$p\left(x\right)=(x-1)\cdot g\left(x\right).....\left(1\right)\Rightarrow g\left(x\right)=\frac{p\left(x\right)}{(x-1)}$

Therefore, $g\left(x\right)$ is obtained by after dividing $p\left(x\right)$ by $(x-1)$ as shown in the above image:

From the division, we get the quotient $g\left(x\right)=x^2-x-6$ and now we factorize it as follows:

$x^2-x-6=x^2-3x+2x-6=x(x-3)+2(x-3)=(x+2)(x-3)$

From equation 1, we get $p\left(x\right)=(x-1)(x+2)(x-3)$.

Hence, $x^3-2x^2-5x+6=(x-1)(x+2)(x-3)$.