Question

If $f\left(x\right)=x^3+x^2-ax+b$ is divisible by $x^2-x$ write the value of a and b.

Answer 1

It is given that the polynomial $f\left(x\right)=x^3+x^2-ax+b$ is divisible by $x^2-x$ which can be rewritten as $x(x-1)$. It means that the given polynomial is divisible by both $x$ and $(x-1)$ that is they both are factors of $f\left(x\right)=x^3+x^2-ax+b$.

Therefore, $x=0$ and $x=1$ are the zeroes of $f\left(x\right)$ that is both $f\left(0\right)=0$ and $f\left(1\right)=0$.

Let us first substitute $x=0$ in $f\left(x\right)=x^3+x^2-ax+b$ as follows:

$f\left(0\right)=0^3+0^2-(a\times 0)+b\Rightarrow 0=0^3+0^2-(a\times 0)+b\Rightarrow 0=0+b\Rightarrow b=0$

Now, substitute $x=1$:

$f\left(1\right)=1^3+1^2-(a\times 1)+b\Rightarrow 0=1+1-a+b\Rightarrow 0=2-a+b\Rightarrow 0=2-a+0(\because b=0)\Rightarrow a=2$

Hence, $a=2$ and $b=0$.