Question

Use the factor theorem to factorize the polynomial $x^3-13x-12$ completely. Hence solve the corresponding polynomial equation.

Answer 1

Step-1 Find the factors of the polynomial $x^3-13x-12$ using factor theorem and hit and trial method:

Let the polynomial be $f\left(x\right)=x^3-13x-12$

Factor theorem, says that if $f\left(x\right)$ is a polynomial of degree $n\ge 1$ and $\text{'}a\text{'}$ is any real number, then, $(x-a)$ is a factor of $f\left(x\right)$, if $f\left(a\right)=0$ .
Putting $x=1$, we get

$\begin{array}{rcl}& \Rightarrow & f\left(1\right)=1^3-13\left(1\right)-12\\ & =& 1-13-12\\ f\left(1\right)& =& -24\ne 0\end{array}$
So, by factor theorem $x=1$ is not a factor of $f\left(x\right)=x^3-13x-12$
Putting $x=-1$, we get

$\begin{array}{rcl}& \Rightarrow & f(-1)=(-1{)}^3-13(-1)-12\\ & =& -1+13-12\\ f(-1)& =& 0\end{array}$
So, by factor theorem $x+1$ is a factor of $f\left(x\right)=x^3-13x-12$

Step- 2 Find the other factor by using long division method:

Divide $f\left(x\right)=x^3-13x-12$ by $(x+1)$
$x+1x^2--x-12 x^3+0x^2-13x-12$
$$\begin{gathered} \frac{x^3+x^2 \\ --}{-x^2-13x-12} \end{gathered}$$
$$\begin{gathered} \frac{-x^2-x \\ ++}{-12x-12} \end{gathered}$$
$$\begin{gathered} \frac{-12x-12 \\ ++}{0} \end{gathered}$$
$\Rightarrow x^3-13x-12=(x+1)(x^2-x-12)$

Step- 3 Find the factors of the obtained quadratic polynomial using middle term factorization:
$\begin{array}{rcl}{x}^2-x-12& =& x^2-4x+3x-12\\ & =& (x-4)(x+3)\end{array}$
So, $x^3-13x-12=(x+1)(x-4)(x+3)$

Step-4 Solve the polynomial equation $f\left(x\right)$:

$f\left(x\right)=0$
$\Rightarrow x^3-13x-12=0$
$\Rightarrow (x+1)(x+3)(x-4)=0$
$\Rightarrow (x+1)=0$ or $(x+3)=0$ or $(x-4)=0$
$\Rightarrow x=-1$or $x=-3$ or $x=4$

Hence, $x^3-13x-12=(x+1)(x-4)(x+3)$ and the solution of the polynomial equation $x^3-13x-12=0$ is $x=-1$or $x=-3$ or $x=4$.