Fita M.

asked • 03/05/20# Fundamental Theorem of Algebra

I need help with these very complex algebra 2 problems. Thank you.

f(x)= x^{4}+21x^{2 }-100

## 2 Answers By Expert Tutors

If you want to factor f(x):

x^{4} + 21x^{2 }-100 =(x^{2 }+ 25)(x^{2 }- 4)

using the perfect square difference formula (a^{2 }- b^{2}) = (a +b)(a-b)

(x^{2 }- 4) = (x + 2)(x - 2)

x^{4} + 21x^{2 }-100=(x^{2 }+ 25)(x^{ }- 2)(x^{ }+ 2)

If you want to find the solution (roots) of f(x):

(x^{2 }+ 25)(x^{ }- 2)(x^{ }+ 2) = 0

(x^{2 }+ 25)=0 or (x^{ }- 2)= 0 or (x +2) =0

The 4 roots are:

x = 5i, x =-5i, x = 2, and x = -2

i is the imaginary = sqr(-1)

Heidi T. answered • 03/05/20

MA in Mathematics, PhD in Physics with 7+ years teaching experience

It isn't at all clear what you are trying to find. I will assume that you are trying to find the roots of this equation.

First, make the substitution that y = x^{2}. This changes your equation to

g(y) = y^{2} + 21y - 100 This is in a form that you can factor easily. You are looking for factors of 100 that have a difference of 21. 25 * 4 = 100, and 25 - 4 - 21, so 25 and 4 are the factors of 100. Factor g(y):

g(y) = (y + 25) (y - 4) ==> y = -25 and y = 4, but y = x^{2}, so this implies that x^{2} = -25 and x^{2} = 4

The second equation gives x = +/- 2; the result from the first depends on whether you want real solutions or can accept imaginary/complex solutions, since there are no real numbers that have a negative square. If imaginary solutions are possible, then x = +/- 5i

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Conner B.

Could you explain what the question is? I see the function f(x), but I don't see what the objective is with the function.03/05/20