Question

find all imaginary solutions of x^4+14x^2-32 by factoring

Answer 1

Cubed will give 2 reals 2 im's

Answer 2

do u know how'd start?

Answer 3

I mean to the forth power

Answer 4

factor the common variable out

Answer 5

Great, what do you get?

Answer 6

We can set
\[f(x)=x^4+14x^2-32\]
When we're finding the zeros, f(x) will be zero because f(x) is y, and when x-intercepts, y will be zero, so
\[f(x)=0\]
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\[x^4+14x^2-32=0\]
\[\left(x^2-2\right) \left(x^2+16\right)=0\]

Answer 7

thank you but it says your suppose to have 4 answers that equal x because the leading term is to the fourth degree

Answer 8

Yes, you will have 4 answers,
\[x^2-2=0~~~~~and~~~~~x^2+16=0 \\ \\ x=\pm \sqrt{2} ~~~~~and~~~~~ x= \pm \sqrt{-16}\]
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\[x=\sqrt2 ~~~~~ x=-\sqrt{2}~~~~~~~~~|~~~~~~~~x=4i~~~~~x=-4i\]

Answer 9

I think that is how you do it. thank you. but did you plug in x^3 an x in the equation to get those answers?