The polynomial $p(x)=x^{4}-2 x^{3}+3 x^{2}-a x+b$ when divided by $(x-1)$ and $(x+1)$ leaves the remainders 5 and 19 respectively. Find the values of $a$ and $b$. Hence, find the remainder when $p(x)$ is divided by $(x-2)$.
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 $b=8]$
$=(16-16+12-10+8)$
$=10$
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