Question

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

Answer 1

Let $f\left(x\right)=x^3-a{x}^2-13x+b$
Because $(x-1)$ and $(x+3)$ are the factors of $f\left(x\right)$
$\therefore f\left(1\right)=0$ and $f(-3)=0$
$f\left(1\right)=0$
$\Rightarrow (1{)}^3-a(1{)}^2-13\left(1\right)+b=0\Rightarrow 1-a-13+b=0$
$\Rightarrow -a+b=12$ ..............(i)
$f(-3)=0$ $\Rightarrow (-3{)}^3-a(-3{)}^2-13(-3)+b=0$
$\Rightarrow -27-9a+39+b=0$
$\Rightarrow -9a+b=-12$ .............(ii)
Subtracting equation (ii) from equation (i)
$(-a+b)-(-9a+b)=12+12$
$\Rightarrow -a+9a=24\Rightarrow 8a=24\Rightarrow a=3$
Put $a=3$ in equation (i) $-3+b=12$
$\Rightarrow b=15$. Hence $a=3$ and $b=15$