# How do you factor completely x^5+8x^3+2x-15?

Jan 13, 2017

This quintic has no factors expressible in terms of elementary functions.

#### Explanation:

Given:

$f \left(x\right) = {x}^{5} + 8 {x}^{3} + 2 x - 15$

If we can find the zeros of $f \left(x\right)$ then we can deduce its factors.

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Magnitude analysis

Note that:

${\textcolor{b l u e}{3}}^{5} = 243 > 237 = 216 + 6 + 15 = 8 \left({\textcolor{b l u e}{3}}^{3}\right) + 2 \left(\textcolor{b l u e}{3}\right) + 15$

Hence we can deduce that any zeros of $f \left(x\right)$ lie inside the circle of radius $3$ in the Complex plane. That is: $\left\mid x \right\mid < 3$

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Descartes' Rule of Signs

The pattern of signs of the coefficients of $f \left(x\right)$ is $+ + + -$. With one change, that means that $f \left(x\right)$ has exactly one positive Real zero.

The pattern of signs of the coefficients of $f \left(- x\right)$ is $- - - -$. With no changes, that means that $f \left(x\right)$ has no negative Real zeros.

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Rational roots theorem

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 15$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 3 , \pm 5 , \pm 15$

We have already eliminated the possibility of $\left\mid x \right\mid \ge 3$ and of $x < 0$, so the only rational possibility is $x = 1$:

$f \left(1\right) = 1 + 8 + 2 - 15 = - 4$

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So $f \left(x\right)$ has no rational zeros. It has an irrational zero in $\left(1 , 3\right)$ and two complex pairs of zeros, all satisfying $\left\mid x \right\mid < 3$.

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$f \left(x\right)$ is a typical quintic, with zeros that are not expressible using elementary functions such as $n$th roots, trigonometric, logarithmic or exponential functions. It is possible to find numerical approximations to the zeros, but there is no sensible algebraic solution.