Question

Find the values 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

$p\left(x\right)=x^3+a{x}^2-13x+b$
as $\left(x-1\right)$ is factor of $p\left(x\right)$
so $p\left(1\right)=0$
$1^3+a\left(1\right)-13\left(1\right)+b=0$
$a+b-12=0$
$a+b=12.......\left(1\right)$
$\therefore x-3$ is factor of $p\left(x\right)$
so, $p\left(3\right)=0$
$3^3+a{\left(3\right)}^2-13\times 3+b=0$
$27+9a-39+b=0$
$9a+b=12.......\left(2\right)$
Solving $\left(1\right)\&\left(2\right)$
$a+b=12$
$\underset{–}{-9a+b=12}$
$\underset{–}{-8a=0}$
$\Rightarrow a=0$
Putting equation $\left(1\right)$
$0+b=12\Rightarrow b=12$