Question

The polynomials $\left(2x^3+x^2-ax+2\right)\mathrm{and}\left(2x^3-3x^2-3x+a\right)$ when divided by (x – 2) leave the same remainder. Find the value of a.

Answer 1

Let $f\left(x\right)=2x^3+x^2-ax+2$ and $g\left(x\right)=2x^3-3x^2-3x+a$.

By remainder theorem, when f(x) is divided by (x – 2), then the remainder = f(2).

Putting x = 2 in f(x), we get

$f\left(2\right)=2\times 2^3+2^2-a\times 2+2=16+4-2a+2=-2a+22$

By remainder theorem, when g(x) is divided by (x – 2), then the remainder = g(2).

Putting x = 2 in g(x), we get

$g\left(2\right)=2\times 2^3-3\times 2^2-3\times 2+a=16-12-6+a=-2+a$

It is given that,

$$\begin{gathered} f\left(2\right)=g\left(2\right) \\ \Rightarrow -2a+22=-2+a \\ \Rightarrow -3a=-24 \\ \Rightarrow a=8 \end{gathered}$$
Thus, the value of a is 8.