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 is a multiple of some positive integer.

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 $g(u) = 0$.

Often, an equation can look difficult to solve, but it can often be quite simple if you can see how to reduce it. A common example, a quartic of the form $ax^4 + bx^2 + c = 0,$ can be made much simpler by the substitution $u = x^2.$ So the equation becomes $au^2+ bu + c = 0,$ a quadratic, which of course is much easier to solve.

For the solution, you can either factor it expicitly $(u-u_1)(u-u_2) = 0,$ where $u_1$ and $u_2$ are the solutions (or roots) of the equation, or you can use the quadratic equation

$$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$

Here are a few examples:

Solve the equation $3x^4+12x^2+7 = 0$.

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Substitute $u = x^2 \implies 3u^2 + 12u + 7 = 0.$

Plugging into the quadratic equation,

$$\begin{aligned} u &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\\ &= \frac{-12 \pm \sqrt{12^2 - 4\cdot 3 \cdot 7}}{2\cdot 3}. \end{aligned}$$

This has two solutions: $u = -0.7, -3.3$.

Finally, $b = \sqrt{u},$ so $b$ has the four purely imaginary roots

$$b = \pm 0.84i,\ \pm 1.81i.\ _\square$$

Solve the equation $x^8+2x^4+1 = 0$.

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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 thus $x^4 = -1$.

As a result, we have $x = \pm \sqrt{\pm i}$. $_\square$

Solve the equation $x^{15}+x^{10}+x^5+1 = 0$.

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Again, we need to identify $n$. It seems $n = 5$ is appropriate here, so we 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 thus $x^{20} = 1$.

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.) $_\square$

• Factorization of Quadratics

• Factorization of Polynomials

• System of Equations - Factorization

• Factorization of Cubics

• Factoring Polynomials with Large Coefficients: Factoring by Extraction