SOLUTION: How do I find the zeros of this polynomial function? f(x) = x<sup>4</sup> + x<sup>3</sup> + 2x<sup>2</sup> + 4x - 8 Thank you very much.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: How do I find the zeros of this polynomial function? f(x) = x<sup>4</sup> + x<sup>3</sup> + 2x<sup>2</sup> + 4x - 8 Thank you very much.      Log On


   

Question 175743This question is from textbook Algebra 2 (McDougal Littel)
: How do I find the zeros of this polynomial function?
f(x) = x4 + x3 + 2x2 + 4x - 8
Thank you very much. This question is from textbook Algebra 2 (McDougal Littel)

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

f(x) = x4 + x3 + 2x2 + 4x - 8

Factor by grouping. 
First rearrange the terms:

f(x) = x4 + 2x2 - 8 + x3 + 4x

Factor the first three terms
Factor x out of the last two

f(x) = (x2 + 4)(x2 - 2) + x(x2 + 4)

Factor out {x2 + 4)

f(x) = (x2 + 4)[(x2 - 2) + x]

Remove parentheses inside bracket

f(x) = (x2 + 4)[x2 - 2 + x]

Rearrange the terms in the bracket,
and change the bracket to parentheses:

f(x) = (x2 + 4)(x2 + x - 2)

Factor the second parentheses:

f(x) = (x2 + 4)(x + 2)(x - 1)

Now we set that = 0

(x2 + 4)(x + 2)(x - 1) = 0

Using the zero factor principle:

x2 + 4 = 0        x + 2 =  0         x - 1 = 0
    x2 = -4           x = -2             x = 1
     x = ±2i

The solutions are 2i, -2i, -2, and 1

Edwin