Question

Find the value of a and b so that the polynomial $x^3-a{x}^2-13x+b$ has $(x-1)$ and $(x+3)$ as factors.

Answer 1

Let $f\left(x\right)=x^3-a{x}^2-13x+b$

Given that, $(x-1)$ and $(x+3)$ are factors of $f\left(x\right)$

According to the factor theorem, if $(x-a)$ is a factor of $f\left(x\right)$, then $f\left(a\right)=0$.

Therefore, $f\left(1\right)=0$ and $f(-3)=0$

$\Rightarrow f\left(1\right)=1-a-13+b=0$

$\Rightarrow b-a=12....\left(i\right)$

And,

$\Rightarrow f(-3)=-27-9a+39+b=0$

$\Rightarrow b-9a=12$

$\therefore b=12+9a$

Substituting the value of $b$ in $\left(i\right)$, we get:

$12+9a-a=12$

$\Rightarrow 8a=0$

$\therefore a=0$

$b=12+9a$

$\therefore b=12+9\left(0\right)=12$

Hence, the value of $a$ is $0$ and that of $b$ is $12$.