Question

Back with a similar question to the last one, but with less data and no real numbers. How do I find the zeros of this equation?
f(x) = x^4 + 4x^2 + 4

Answer 1

solve the quadratic equation in \(x^2\) or factor as \((x^2+2)^2\)
the zeros are complex

Answer 2

Could you explain this a little more please? It's supposed to be a review, but I haven't done stuff like this for over a year and the book is super vague because the writer expected us to know how to do all of this.

Answer 3

\[f(x)=(x^2+2)^2+4\]
0 = x^2 + 2
-2 = x^2
2i = x & -2i = x

Answer 4

Basically, you can factor it like a quadratic because the first power of x is twice the second power of x and the 3rd term is a number. So you factor it the same way you would factor
\[x^{2} + 4x + 4\]

Answer 5

(x+2)(x+2)?
wut.

Answer 6

Well, yes, that would be the way you factor it. But what satelite was saying is that you just need to make it (x^2+2)(x^2+2). The reason its x^2 is because that is the power that is doubled in the first term. So for example, if we had x^6 + x^3 + 1, the factorization would be (x^3 + __)(x^3 + ___) It would be factored the same way, just the power of the variable after you factor would be different. So yes, its + 2 + 2, but with x^2 instead, so you have (x^2+2)(x^2+2).

Answer 7

\[(x^2+2)(x^2+2)=x^4+2x^2+2x^2+4\]
\[=x^4+4x+4\]
So does this mean that my zeros are art: x=i-2 and x=i+2?

Answer 8

are at*

Answer 9

Close. And sorry for the delay. Well, you would set your factors equal to 0, so you would have
x^2 + 2 = 0
x^2 = -2
square root both sides
x = +/- sqrt(-2)
x = +/- sqrt(2)*sqrt(-1) =
\[x = \pm i \sqrt{2} \]