# Can y=x^5 - x^3 + x - 2 x - 5 be factored? If so what are the factors ?

May 1, 2016

Not algebraically.

#### Explanation:

I think there's at least one typo in the question, so consider three polynomials:

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

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

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

By the rational root theorem, the only rational zeros of these polynomials are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 5$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 5$

None of these are zeros of ${f}_{1} \left(x\right)$ or ${f}_{2} \left(x\right)$, so these polynomials have no factors with rational coefficients.

Though they each have one Real zero and two pairs of Complex zeros there are no nice algebraic solutions to either.

In the case of ${f}_{3} \left(x\right)$ we find ${f}_{3} \left(1\right) = 0$, so $\left(x - 1\right)$ is a factor:

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

The remaining quartic factor does have an algebraic solution, but it is not at all nice.