Question

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

Answer 1

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

Explanation:

Given:

`f(x) = x^5+8x^3+2x-15`

If we can find the zeros of `f(x)` then we can deduce its factors.

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Magnitude analysis

Note that:

`color(blue)(3)^5 = 243 > 237 = 216+6+15 = 8(color(blue)(3)^3)+2(color(blue)(3))+15`

Hence we can deduce that any zeros of `f(x)` lie inside the circle of radius `3` in the Complex plane. That is: `abs(x) < 3`

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Descartes' Rule of Signs

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

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

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Rational roots theorem

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `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:

`+-1, +-3, +-5, +-15`

We have already eliminated the possibility of `abs(x) >= 3` and of `x < 0`, so the only rational possibility is `x=1`:

`f(1) = 1+8+2-15 = -4`

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So `f(x)` has no rational zeros. It has an irrational zero in `(1, 3)` and two complex pairs of zeros, all satisfying `abs(x) < 3`.

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`f(x)` 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.