Question

Find the value of $a$ and $b$ in the polynomial $f\left(x\right)=2x^3+a{x}^2+bx+10$, if it is exactly divisible by $x+2$ and $2x-1$.​

Answer 1

Given: $f\left(x\right)=2x^3+a{x}^2+bx+10$
A polynomial $f\left(x\right)$ is exactly divisible by $x-a$ if and only if $f\left(a\right)=0$.

$\text{f(x)}$ is exactly divisible by $\text{x+2}$.

$\therefore$$\text{f(x)}=0$
$2(-2{)}^3+\text{a}(-2{)}^2+\text{b}(-2)+10=0-16+4\text{a}-2\text{b}+10=0$
$4\text{a}-2\text{b}=6$
$2\text{a}-\text{b}=3$
$\text{b}=2\text{a}-3$-(i)

Also, $\text{f(x)}$ is exactly divisible by $(2\text{x}-1)$.
$\therefore \text{f}\left(\begin{array}{c}\frac{1}{2}\end{array}\right)$
$2{\left(\begin{array}{c}\frac{1}{2}\end{array}\right)}^3+\text{a}{\left(\begin{array}{c}\frac{1}{2}\end{array}\right)}^2+\text{b}\left(\begin{array}{c}\frac{1}{2}\end{array}\right))+10=0$

$\frac{1}{4}+\frac{\text{a}}{4}+\frac{\text{b}}{2}+10=0$

$1+\text{a}+2\text{b}+40=0$
$\text{a}+2\text{b}=-41$-(ii)

Step 2: Find the value of a and b

Put the value of b in equation (ii)
$\text{a}+2(2\text{a}-3)=-41$
$\text{a}+4\text{a}-6=-41$
$5\text{a}=-35$
$\text{a}=-7$

Put the value of a in equation (i)
$\text{b}=2(-7)-3=-14-3$
$\text{b}=-17$

Hence, $\text{a}=-7$ and $\text{b}=-17$.