Question

Let $f\left(x\right)=x^3+a{x}^2+bx+c.$ Given $f\left(1\right)=4,f\left(2\right)=8.$ Then the value of $f\left(6\right)-f(-3)$ is equal to

• A. $0$
• B. $6$
• C. $36$
• D. $216$

Answer 1

The correct option is D $216$

$f\left(1\right)=4$
This means when $f\left(x\right)$ is divided by $x-1,$ the remainder is $4=1\times 4.$
$f\left(2\right)=8$
This means when $f\left(x\right)$ is divided by $x-2,$ the remainder is $8=2\times 4.$
So, we can write
$f\left(x\right)=(x-1)(x-2)(x-\alpha )+4x$
where $\alpha$ is a root of $f\left(x\right)=0$

Then $f\left(6\right)=20(6-\alpha )+24\cdots \left(1\right)$
$f(-3)=20(-3-\alpha )-12\cdots \left(2\right)$
Subtracting $\left(1\right)$ and $\left(2\right),$ we get
$f\left(6\right)-f(-3)=216$