# Factoring Compound Quadratics - \(ax^4 + bx^2 + c\)

A compound quadratic is a polynomial that can be expressed in the form \(ax^{2n}+bx^{n}+c\), where \(a \neq 0,b\) and \(c\) are constants and \(n\) is a positive integer. This can be generalized to compound polynomials, where the degree of the terms are a multiple of some positive integer.

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## Method

To solve a compound quadratic equation such as \(ax^{2n}+bx^{n}+c\), one introduces a new variable \(u = x^n\). The equation becomes \(au^2+bu+c = 0\), and we can then solve for \(u\), after which we may solve for \(x\).

In general, it is possible to solve any compound polynomial equation \(f(x) = 0 \), by considering it as a polynomial \( g(u) \) with the substitution \(u = x^n\), if one knows the solutions to \(g(u) = 0\).

## Worked Examples

Solve the equation \(x^8+2x^4+1 = 0\).

First, we need to identify \(n\). Since the LHS is a polynomial in \(x\) as well as \(x^4\), it seems suitable that we choose \(n = 4\). We thus substitute \(u = x^4\), obtaining \(u^2+2u+1 = 0\).

Factoring this equation, we get that \(u^2+2u+1 = (u+1)^2 = 0\); hence, \(u+1 = 0\) and so \(x^4 = -1\).

As a result, we have \(x = \pm \sqrt{\pm i}\).

Solve the equation \(x^{15}+x^{10}+x^5+1 = 0\).

Again, we need to identify \(n\). It seems \(n = 5\) is appropriate here, so we thus substitute \(u = x^5\), obtaining \(u^3+u^2+u+1 = 0\).

We multiply both sides by \(u-1\), and reject \(u = 1\). We therefore get \((u-1)(u^3+u^2+u+1) = u^4-1 = 0\); hence, \(u^4 = 1\) and so \(x^{20} = 1\).

As a result, we have \(x = e^{k\pi / 10}\), with \(k \in \mathbb{Z}\) and \(4 \nmid k\). (To see how one arrives at the solution set, look up the roots of unity.)

**Cite as:**Factoring Compound Quadratics - \(ax^4 + bx^2 + c\).

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/factor-polynomials-ax4-bx2-c/