Question

Factorise : $x^3+13x^2+31x-45$ by factor theorem.

Answer 1

Let the given polynomial be $p\left(x\right)=x^3+13x^2+31x-45$.

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

$Forx=0p\left(0\right)=(0{)}^3+13(0{)}^2+31\times \left(0\right)-45=0+0+0-45=-45\ne 0\therefore p\left(0\right)\ne 0$

$Forx=1p\left(1\right)=(1{)}^3+13(1{)}^2+31\times \left(1\right)-45=1+13+31-45=45-45=0\therefore p\left(1\right)=0$

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 dividing $p\left(x\right)$ by $(x-1)$ as shown in the above image:

From the division, we get the quotient $g\left(x\right)=x^2+14x+45$ and now we factorise it as follows:

$x^2+14x+45=x^2+9x+5x+45=x(x+9)+5(x+9)=(x+5)(x+9)$

From equation 1, we get $p\left(x\right)=(x-1)(x+5)(x+9)$.

Hence, $x^3+13x^2+31x-45=(x-1)(x+5)(x+9)$.