Question

When a polynomial $2x^3+3x^2+ax+b$ is divided by $(x-2)$ leaves remainder $2$, and $(x+2)$ leaves remainder $-2$. Find $a$ and $b$.

Answer 1

The given polynomial is $P\left(x\right)=2x^3+3x^2+ax+b$

It is also given that if $P\left(x\right)$ is divided by $(x-2)$ then it will leave the remainder $2$ and if divided by $(x+2)$ then the remainder will be $-2$ which means that $P\left(2\right)=2$ and $P(-2)=-2$ (Using Remainder theorem).

Let us first substitute $x=2$ in $P\left(x\right)=2x^3+3x^2+ax+b$ as shown below:

$\Rightarrow P\left(2\right)=2(2{)}^3+3(2{)}^2+(a\times 2)+b\Rightarrow 2=(2\times 8)+(3\times 4)+2a+b\Rightarrow 2=16+12+2a+b$

$\Rightarrow 2=28+2a+b\Rightarrow 2a+b=2-28\Rightarrow 2a+b=-26.........\left(1\right)$

Now, substitute $x=-2$ in $P\left(x\right)=2x^3+3x^2+ax+b$ as shown below:

$\Rightarrow P(-2)=2(-2{)}^3+3(-2{)}^2+(a\times -2)+b\Rightarrow -2=(2\times -8)+(3\times 4)-2a+b\Rightarrow -2=-16+12-2a+b$

$\Rightarrow -2=-4-2a+b\Rightarrow -2a+b=-2+4\Rightarrow -2a+b=2.........\left(2\right)$

Now adding equations $\left(1\right)$ and $\left(2\right),$ we get

$(2a-2a)+(b+b)=-26+2\Rightarrow 2b=-24\Rightarrow b=-\frac{24}{2}\Rightarrow b=-12$

Now substitute the value of ${}^{\prime }{b}^{\prime }$ in equation $2$:

$-2a+(-12)=2\Rightarrow -2a-12=2\Rightarrow -2a=2+12\Rightarrow -2a=14\Rightarrow a=-\frac{14}{2}\Rightarrow a=-7$

Hence $a=-7$ and $b=-12$