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 remainder are respectively $5$ and $19$. Determine the remainder when $f\left(x\right)$ is divided by $(x-2)$.

Answer 1

When f(x) is divided by x-1 and x+1 the remainder are 5 and 19 respectively.
$\therefore f\left(1\right)=5$ and $f(-1)=19$
$\Rightarrow (1{)}^4-2\times (1{)}^3+3\times (1{)}^2-a\times 1+b=5$
and $(-1{)}^4-2\times (-1{)}^3+3\times (-1{)}^2-a\times (-1)+b=19$
$\Rightarrow 1-2+3-a+b=5$
and $1+2+3+a+b=19$
$\Rightarrow 2-a+b=5$ and $6+a+b=19$
$\Rightarrow -a+b=3$ and $a+b=13$
Adding these two equations, we get
$(-a+b)+(a+b)=3+13$
$\Rightarrow 2b=16\Rightarrow b=8$
Putting $b=8$ and $-a+b=3$, we get
$-a+8=3\Rightarrow a=-5\Rightarrow a=5$
Putting the values of a and b in
$f\left(x\right)=x^4-2x^3+3x^2-5x+8$
The remainder when f(x) is divided by (x-2) is equal to f(2).
So, Remainder $=f\left(2\right)=(2{)}^4-2\times (2{)}^3+3\times (2{)}^2-5\times 2+8=16-16+12-10+8=10$