Question

What number should be added to $\text{f(x)}=12{\text{x}}^3+16\text{x}2-5\text{x}-8$ so that the resulting polynomial is exactly divisible by $(2\text{x}-1)$.​

Answer 1

Step: Find the number

Let's assume the required number to be k.
$\text{f(x)}=12{\text{x}}^3+16{\text{x}}^2-5\text{x}-8+\text{k}$

A polynomial $\text{f(x)}$ is exactly divisible by $\text{(x-a)}$ if and only if $\text{f(a)}=0$.

$\text{f(a)}=0$ is exactly divisible by $\text{(2x-1)}=0$.
$\text{2x-1}=0\Rightarrow \text{x}=\frac{1}{2}$
$\therefore \text{f}\left(\begin{array}{c}\frac{1}{2}\end{array}\right)=0$
$12{\left(\begin{array}{c}\frac{1}{2}\end{array}\right)}^3-16{\left(\begin{array}{c}\frac{1}{2}\end{array}\right)}^2-\frac{5}{2}-8+\text{k}=0$
$12\left(\begin{array}{c}\frac{1}{8}\end{array}\right))+16\left(\begin{array}{c}\frac{1}{4}\end{array}\right)-\frac{5}{2}-8+\text{k}=0$
$\frac{3}{2}+4-\frac{5}{2}-8+\text{k}=0$
$-1-4+\text{k}=0$
$\text{k}=5$

Hence, $5$ should be added to $12{\text{x}}^3+16{\text{x}}^2-5\text{x}-8$.