How do you factor #x^5 - 5x^4 - x^3 + x^2 + 4 = 0#?

1 Answer
Dec 27, 2015

Answer:

#x^5-5x^4-x^3+x^2+4 = (x-1)(x+1)(x^3-5x^2-4)#

There are no simpler factors with rational coefficients.

Explanation:

Let #f(x) = x^5-5x^4-x^3+x^2+4#

First note that the sum of the coefficients is zero.

That is #1-5-1+1+4 = 0#

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

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

#-1-5+1+1+4 = 0#

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

#(x-1)(x+1) = x^2-1#, 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)#

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