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, the remainder when $f\left(x\right)$ is divisible by $(x-2)$ is

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

The correct option is D $10$

When $x^4-{2x}^3+3x^2-ax+b$ is divide by $x-1,$ remainder is $5.$
So, substituting for $x$ is $1,$ in the above, we get

$5=1-2+3-a+b$
$\therefore -a+b=3$----(1)
When $x^4-{2x}^3+3x^2-ax+b$ is divide by $x+1$, remainder is $19.$
So, substituting for $x$ is $-1,$ in the above, we get

$19=1+2+3+a+b$
$a+b=13$----(2)
Solving (1) and (2), we get $a=5,b=8$

So polynomial becomes $x^4-{2x}^3+3x^2-5x+8$
The remainder when $x^4-{2x}^3+3x^2-5x+8$ is divided by $x-2$ is by plugging in $x$ as $2$ in the given polynomial, we get

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

so remainder is $10$
So, option D.