Question

a student has factored the polynomial equation $f(x)=x^{4}-8x^{3}+26x^{2}-48x + 45$ to $f(x)=(x - 3)(x - 3)(x^{2}-2x + 5)$. determine the roots of the polynomial equation.
$x={-3,1pm2i}
$ $x = {3,2pm i}
$ because the factored form has two duplicate factors of $(x - 3)$, they will have the same root come from them. in order to solve $x^{2}-2x + 5=0$, you will have to use the quadratic formula.
$x={-3,2pm i}
$ $x={3,1pm2i}$

Explanation:

Step1: Set each factor equal to zero

For the factor $(x - 3)$, set $x-3=0$, then $x = 3$. Since $(x - 3)$ appears twice, $x = 3$ is a root with multiplicity 2.

Step2: Solve the quadratic factor

For the quadratic factor $x^{2}-2x + 5=0$, use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 1$, $b=-2$, and $c = 5$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-2)^{2}-4\times1\times5=4 - 20=-16$. Then $x=\frac{2\pm\sqrt{-16}}{2}=\frac{2\pm4i}{2}=1\pm2i$.

Answer:

$x=\{3,1\pm2i\}$