Question

Show that $(2x+7)$ is a factor of $2x^3+7x^2-4x-14.$ Hence factorize $2x^3+7x^2-4x-14$.

Answer 1

Step- 1 Use the factor theorem:

If $f\left(x\right)$ is a polynomial of $n\ge 1$and $a$ is any real number, then $x-a$ is a factor of $f\left(x\right)$ if $f\left(a\right)=0$.

Let $f\left(x\right)=2x^3+7x^2-4x-14$.

If $(2x+7)$ is a factor of $f\left(x\right)=2x^3+7x^2-4x-14$, then $f\left(-\frac{7}{2}\right)$ will be $0$.

$f\left(-\frac{7}{2}\right)$$=2{(-\frac{7}{2})}^3+7{(-\frac{7}{2})}^2-4(-\frac{7}{2})-14$

$=-\frac{343}{4}+\frac{343}{4}+14-14$
$=0$

Thus, $(2x+7)$ is a factor of $f\left(x\right)=2x^3+7x^2-4x-14$.

Step- 2 Find the other factor by long division:

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

Clearly we got, $2x^3+7x^2-4x-14=(2x+7)(x^2-2)$.

Also, $x^2-2=(x+\sqrt{2})(x-\sqrt{2})$. Thus, $2x^3+7x^2-4x-14=(2x+7)(x-\sqrt{2})(x-\sqrt{2})$.

Hence, $2x+7$ is a factor of $2x^3+7x^2-4x-14.$and factorization of $2x^3+7x^2-4x-14=(2x+7)(x-\sqrt{2})(x-\sqrt{2})$.