How do you factor x^5 - 5x^4 - x^3 + x^2 + 4 = 0x5−5x4−x3+x2+4=0?

1 Answer
George C.
Dec 27, 2015

x^5-5x^4-x^3+x^2+4 = (x-1)(x+1)(x^3-5x^2-4)x5−5x4−x3+x2+4=(x−1)(x+1)(x3−5x2−4)

There are no simpler factors with rational coefficients.

Explanation:

Let f(x) = x^5-5x^4-x^3+x^2+4f(x)=x5−5x4−x3+x2+4

First note that the sum of the coefficients is zero.

That is 1-5-1+1+4 = 01−5−1+1+4=0

So f(1) = 0f(1)=0 and (x-1)(x−1) is a factor

Reversing the signs on the coefficients of odd degree we also find:

-1-5+1+1+4 = 0−1−5+1+1+4=0

So f(-1) = 0f(−1)=0 and (x+1)(x+1) is a factor

(x-1)(x+1) = x^2-1(x−1)(x+1)=x2−1, so let's separate out this factor to find:

x^5-5x^4-x^3+x^2+4 = (x^2-1)(x^3-5x^2-4)x5−5x4−x3+x2+4=(x2−1)(x3−5x2−4)

By the rational root theorem, the only possible rational zeros of the remaining cubic factor (x^3-5x^2-4)(x3−5x2−4) are +-1±1, +-2±2, +-4±4.

None of these work, so this cubic only has factors with irrational and/or Complex coefficients.

If this question was asked in Precalculus, I might offer a solution of the cubic, but it's probably beyond the scope of your Algebra course, so let's leave the factoring at this point.

Related questions
See all questions in Factoring Completely
Impact of this question
1258 views around the world
You can reuse this answer
Creative Commons License