Question

If $(x-2)$ is a factor of the expression $2x^3+a{x}^2+bx-14$ and when the expression is divided by $(x-3)$, it leaves a remainder $52$, find the valves of $a$ and $b$.

Answer 1

Given: $(x-2)$ is a factor of $2x^3+a{x}^2+bx-14=0$

$x=2$ will be a root of the eexpression

$\Rightarrow 2(2{)}^3+a(2{)}^2+b\left(2\right)-14=0$

$\Rightarrow 16+4a+2b-14=0$

$\Rightarrow 4a+2b+2=0$

$\Rightarrow 4a+2b=-2$

$\Rightarrow 2a+b=-1$ $...\left(i\right)$

When then the expression is divided by $(x-3),$ it leaves a remainder $52$

$\therefore 2x^3+a{x}^2+bx-14=52$ when $x=3$

Let us substitute $x=3$ in the expression

$2(3{)}^3+a(3{)}^2+b\left(3\right)-14=52$

$2(3{)}^3+a(3{)}^2+b\left(3\right)-66=0$

$54+9a+3b-66=0$

$9a+3b=12$

$3a+b=4$ $...\left(ii\right)$

Solving equation $\left(i\right)$ and $\left(ii\right)$,

$2a+b=-1$
$3a+b=4$
$(-)$ $(-)$ $(+)$
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$-a=-5$

$\Rightarrow a=5$

Substitute $a=5$ in $\left(ii\right)$

$\Rightarrow 3\times 5+b=4$

$\Rightarrow b=4-15=-11$

$\therefore a=5$ and $b=-11$