Question

Show that $2x+7$ is a factor of $2x^3+5x^2-11x-14$. Hence, factorize the given expression completely, using the factor theorem.

Answer 1

Step-1 Check $2x+7$ is a factor of $2x^3+5x^2-11x-14$ or not by using the factor theorem:

Let, $f\left(x\right)=2x^3+5x^2-11x-14,g\left(x\right)=2x+7$.

Factor theorem states that a polynomial $f\left(x\right)$ has a factor $\left(x-a\right)$, if and only if, $f\left(a\right)=0$.

According to factor theorem, $f\left(x\right)$ has a factor $\left(x+\frac{7}{2}\right)$ if and only if $f\left(-\frac{7}{2}\right)=0$.

$\begin{array}{rcl}& \Rightarrow & f\left(-\frac{7}{2}\right)=2{\left(-\frac{7}{2}\right)}^3+5{\left(-\frac{7}{2}\right)}^2-11\left(-\frac{7}{2}\right)-14\\ & =& -\frac{343}{4}+\frac{245}{4}+\frac{77}{2}-14\\ & =& \frac{-399+399}{4}\\ & =& 0\end{array}$

Hence, $\left(2x+7\right)$ is a factor of the given polynomial.

Step-2 Divide $f\left(x\right)=2x^3+5x^2-11x-14$ by $\left(2x+7\right)$using the long division method:
$2x+7x^2-x-22x^3+5x^2-11x-14$
$$\begin{gathered} \frac{2x^3+7x^2 \\ --}{-2x^2-11x} \end{gathered}$$
$$\begin{gathered} \frac{-2x^2-7x \\ ++}{-4x-14} \end{gathered}$$
$$\begin{gathered} \frac{-4x-14 \\ ++}{0} \end{gathered}$$

Therefore, $f\left(x\right)=\left(2x+7\right)\left(x^2-x-2\right)$.

Step-3 Factorize the given expression $f\left(x\right)=\left(2x+7\right)\left(x^2-x-2\right)$:

$\begin{array}{rcl}f\left(x\right)& =& \left(2x+7\right)\left(x^2-x-2\right)\\ & =& \left(2x+7\right)\left(x^2-x-2\right)\\ & =& \left(2x+7\right)\left(x^2-2x+x-2\right)\\ & =& \left(2x+7\right)\left(x(x-2)+1(x-2)\right)\\ & =& \left(2x+7\right)(x+1)(x-2)\end{array}$

Hence, $2x+7$ is a factor of $2x^3+5x^2-11x-14$. Therefore, factors of the polynomial $2x^3+5x^2-11x-14$ is $\left(2x+7\right)(x+1)\left(2x+7\right)$.