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

Dec 27, 2015

${x}^{5} - 5 {x}^{4} - {x}^{3} + {x}^{2} + 4 = \left(x - 1\right) \left(x + 1\right) \left({x}^{3} - 5 {x}^{2} - 4\right)$

There are no simpler factors with rational coefficients.

Explanation:

Let $f \left(x\right) = {x}^{5} - 5 {x}^{4} - {x}^{3} + {x}^{2} + 4$

First note that the sum of the coefficients is zero.

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

So $f \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor

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

$- 1 - 5 + 1 + 1 + 4 = 0$

So $f \left(- 1\right) = 0$ and $\left(x + 1\right)$ is a factor

$\left(x - 1\right) \left(x + 1\right) = {x}^{2} - 1$, so let's separate out this factor to find:

${x}^{5} - 5 {x}^{4} - {x}^{3} + {x}^{2} + 4 = \left({x}^{2} - 1\right) \left({x}^{3} - 5 {x}^{2} - 4\right)$

By the rational root theorem, the only possible rational zeros of the remaining cubic factor $\left({x}^{3} - 5 {x}^{2} - 4\right)$ are $\pm 1$, $\pm 2$, $\pm 4$.

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.