Can someone please help me figure this out? I don't think

>>376545
the second to last line is correct, I don't understand the others.

Basically you need to substitute z = x^2 and you get

z^2 + 4 z - 21 = 0

which can be solved as usual (either by looking or with the formula
z1 = -4/2 + sqrt(2^2 + 21) = -2 + 5 = 3
z2 = -4/2 - sqrt(2^2 + 21) = - 7

so you get
(*) z^2 + 4 z - 21 = (z - 3)(z + 7)
and by substituting back
x^4 + 4x^2 - 21 = (x^2 - 3)(x^2 + 7) = 0

The first factor gives two solutions, x = +-sqrt(3), the second no real solution since x^2 + 7 > 0.

You get always four complex zeros / linear factors (counting multiplicity) but at most four real zeros.

An easy example: f(x) = x^2 + a
f(x) = 0 has no solution if a > 0, one solution if a = 0, namely 0 (double), and two solutions if a < 0, namely +-sqrt(|a|).

File: 20170905_110347.jpg (2MB, 2660x1824px) Image search: [Google]

2MB, 2660x1824px

Okay thank you anon, I think I have that one sort of figured out. Is sqrt(7)i an answer? I have down sqrt(3), sqrt(3)i, and sqrt(7)i, but is it correct to put i? I will probably need to wait and get some feedback from my teacher to fully understand.

I also have another problem that i don't even know where to start, if you or anyone wanted to help with this too. I imagine it has a similar process. It is pic related

>>376593
+sqrt(3), -sqrt(3), +sqrt(7)i, -sqrt(7)i

For such functions: find a zero by guessing, use divisors of 6, and you will find that 2 is a zero

2^3 - 2(2^2) -3(2) + 6 = 0

so you know that (x - 2) is a linear factor.

>>376606
Man, I'm confused when to use negative and when to use i. I think that's what is throwing me off.

And thank you for that tip, that is helpful. I think I figured that problem out by factoring. My solution set was 2, sqrt(3), and sqrt(3)i. Or maybe it's supposed to be -sqrt(3)...?

>>376615
the last one is -sqrt(3) so three real zeros.
(x-2)(x^2 - 3) = (x - 2)(x - sqrt(3))(x + sqrt(3)).

Note that you can't have only one complex zero c if the coefficients of the polynome are real numbers:
(x - c)(x - r1)(x - r2)...(x - r) = x^n - (c + r1+ r2 + ... + r)x^(n-1)+... and the second coefficient must be real.