Question

Hardly a Pre-Calc question, more Alegebra 2.
What are the zeros of the function x^6-x^4-16x^2+16.
Pretty sure I have the answer but can you help me through the steps of factoring by grouping?

Answer 1

what do you have so far?

Answer 2

I think the zeros are (x+1)(x-1) (x+2) (x-2) (x+i) (x-i)

Answer 3

factor by grouping to get


x^6-x^4-16x^2+16

(x^6-x^4)+(-16x^2+16)

x^4(x^2-1)-16(x^2-1)

(x^4-16)(x^2-1)

(x^2-4)(x^2+4)(x^2-1)

(x-2)(x+2)(x^2+4)(x^2-1)

Answer 4

oh sry, factor x^2 - 1 to get (x-1)(x+1), so the whole thing factors to

(x-2)(x+2)(x^2+4)(x-1)(x+1)

Answer 5

So there are no imaginary zeros?

Answer 6

yes there are

Answer 7

I dont understand, what the imaginary zeros are, was i right the first time?

Answer 8

you were close

Answer 9

solve x^2+4 = 0

Answer 10

x^2=-4

Answer 11

umm

Answer 12

confused with how i square root the -4

Answer 13

break it up like this

\[\Large \sqrt{-4} = \sqrt{-1*4}\]

\[\Large \sqrt{-4} = \sqrt{-1}*\sqrt{4}\]

Answer 14

ahh ok, so
x=+-2i

Answer 15

good

Answer 16

to get the other real roots, set them equal to 0 and solve for x

Answer 17

I understand now! Thank you very much for your help! :D

Answer 18

yw

Answer 19

Going to close it, farewell friend.

Answer 20

btw might I add that you can also solve problem easily by using synthetic division