Question

Use remainder theorem to factorize the polynomial $2x^3+3x^2-9x-10$.

Answer 1

Step-1 Find the possible factor of a given polynomial using the hit and trial method:

Let the polynomial $f\left(x\right)=2x^3+3x^2-9x-10$.

For $x=1$:

$\begin{array}{rcl}f\left(1\right)& =& 2{\left(1\right)}^3+3{\left(1\right)}^2-9\left(1\right)-10\\ & =& 2+3-9-10\\ & =& -14\\ & \ne & 0\end{array}$

So, $x-1$ is not a factor of $f\left(x\right)=2x^3+3x^2-9x-10$.

For $x=-1$:

$\begin{array}{rcl}f(-1)& =& 2{(-1)}^3+3{(-1)}^2-9(-1)-10\\ & =& -2+3+9-10\\ & =& 0\end{array}$

So, $x+1$ is a factor of $f\left(x\right)=2x^3+3x^2-9x-10$.

For $x=2$:

$\begin{array}{rcl}f\left(2\right)& =& 2{\left(2\right)}^3+3{\left(2\right)}^2-9\left(2\right)-10\\ & =& 16+12-18-10\\ & =& 0\end{array}$

So, $x-2$ is a factor of $f\left(x\right)=2x^3+3x^2-9x-10$.

Therefore, product of factors is $\left(x-2\right)\left(x+1\right)=x^2-x-2$.

Step-2 Find the other factor to divide $f\left(x\right)=2x^3+3x^2-9x-10$ by $x^2-x-2$ by using long division method:
$x^2-x-22x+52x^3+3x^2-9x-10$
$$\begin{gathered} \frac{2x^3-2x^2-4x \\ -++}{5x^2-5x-10} \end{gathered}$$
$$\begin{gathered} \frac{5x^2-5x-10 \\ -++}{0} \end{gathered}$$

Hence, $\left(x-1\right),\left(x-2\right)\text{and}\left(2x+5\right)$ are factors of the given polynomial $2x^3+3x^2-9x-10$.