Question

I am very confused on how to get i, how do you do this problem?

Answer 1

given polynomial is,

$(x-2)(x^2-12x+40)=0_{}$

let us find thefactors of the polynomial,

$\begin{gathered} x-2=0 \\ x=2 \\ \\ \end{gathered}$

for the polynomial,

$x^2-12x+40=0$

use the formula to find the factors,

$\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

where,

$\begin{gathered} a=1 \\ b=-12 \\ c=40 \end{gathered}$

thus,

$\begin{gathered} =\frac{-\mleft(-12\mright)\pm\sqrt[]{(-12)^2-4\times1\times40}}{2\times1} \\ =\frac{12\pm\sqrt[]{144-160}}{2} \\ =\frac{12\pm\sqrt[]{-16}}{2} \\ =\frac{12\pm\sqrt[]{2\times8}}{2} \\ =\frac{12\pm4i}{2} \\ =\frac{2(6\pm2i)}{2} \\ =6\pm2i \end{gathered}$

thus,

$\begin{gathered} x=6+2i \\ x=6-2i \end{gathered}$

the factors are,

$\begin{gathered} x=2 \\ x=6+2i \\ x=6-2i \end{gathered}$