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*1*40}}{2*1} \\ =\frac{12\pm\sqrt[]{144-160}}{2} \\ =\frac{12\pm\sqrt[]{-16}}{2} \\ =\frac{12\pm\sqrt[]{2*8}}{2} \\ =(12\pm4i)/(2) \\ =(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}`