The polynomial p(x)

Solution:

Let

$p(x)=x^{4}-2 x^{3}+3 x^{2}-a x+b$

Now,

When $p(x)$ is divided by $(x-1)$, the remainder is $p(1)$.

When $p(x)$ is divided by $(x+1)$, the remainder is $p(-1)$

Thus, we have:

$p(1)=\left(1^{4}-2 \times 1^{3}+3 \times 1^{2}-a \times 1+b\right)$

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

$=2-a+b$

And,

$p(-1)=\left[(-1)^{4}-2 \times(-1)^{3}+3 \times(-1)^{2}-a \times(-1)+b\right]$

$=(1+2+3+a+b)$

$=6+a+b$

Now,

$2-a+b=5 \quad \ldots(1)$

$6+a+b=19 \ldots(2)$

Adding $(1)$ and $(2)$, we get:

$8+2 b=24$

$\Rightarrow 2 b=16$

$\Rightarrow b=8$

By putting the value of b, we get the value of a, i.e., 5.
∴ a = 5 and b = 8

Now,

$f(x)=x^{4}-2 x^{3}+3 x^{2}-5 x+8$

Also,

When $p(x)$ is divided by $(x-2)$, the remainder is $p(2)$.

Thus, we have:

$p(2)=\left(2^{4}-2 \times 2^{3}+3 \times 2^{2}-5 \times 2+8\right) \quad[a=5$ and $b=8]$

$=(16-16+12-10+8)$

$=10$