Question:
The polynomials $\left(2 x^{3}+x^{2}-a x+2\right)$ and $\left(2 x^{3}-3 x^{2}-3 x+a\right)$ when divided by $(x-2)$ leave the same remainder. Find the value of $a$.
Solution:
Let $f(x)=2 x^{3}+x^{2}-a x+2$ and $g(x)=2 x^{3}-3 x^{2}-3 x+a$
By remainder theorem, when $f(x)$ is divided by $(x-2)$, then the remainder $=f(2)$.
Putting x = 2 in f(x), we get
$f(2)=2 \times 2^{3}+2^{2}-a \times 2+2=16+4-2 a+2=-2 a+22$
By remainder theorem, when g(x) is divided by (x – 2), then the remainder = g(2).
Putting x = 2 in g(x), we get
$g(2)=2 \times 2^{3}-3 \times 2^{2}-3 \times 2+a=16-12-6+a=-2+a$
It is given that,
$f(2)=g(2)$
$\Rightarrow-2 a+22=-2+a$
$\Rightarrow-3 a=-24$
$\Rightarrow a=8$
Thus, the value of a is 8.