Question

If $f\left(x\right)=x^4-2x^3+3x^2-ax+b$ is a polynomial such that when it is divided by $x-1$ and $x+1$, the remainders are $5$ and $19$, respectively. Determine the remainder when $f\left(x\right)$ is divided by $x-2$.

Answer 1

When $f\left(x\right)$ is divided by $x-1$ and $x+1$, the remainders are 5 and 19 respectively.

Therefore, $f\left(1\right)=5$ and $f(-1)=19$

$1-2+3-a+b=5$ and $1+2+3+a+b=19$

$-a+b=3$ and $a+b=13$

Adding these two, we get,

$b=8$

Therefore, $a=5$

Substituting these values of a and b in f(x), we get,

$f\left(x\right)=x^4-2x^3+3x^2-5x+8$

The remainder when $f\left(x\right)$ is divided by $x-2$ is equal to $f\left(2\right)$.

Therefore,

Remainder = $f\left(2\right)=2^4-2\times 2^3+3\times 2^2-5\times 2+8=10$