Tutorial: Solving polynomial equations

Adaptive game version

Go to Part A: Solving linear and quadratic equations

This tutorial: Part B: Solving cubic and higher order polynomial equations

(This topic is also in Section 0.5 in Finite Mathematics and Applied Calculus)

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Basics

Recall the following from Part A of this tutorial.

Polynomial and polynomial equation
A polynomial is an algebraic expression of the form

$ax^n + bx^{n-1} + \cdots + rx + s$where $a, b, \dots r$ and $s$ are constants, called the coefficients of the polynomial. The largest exponent of $x$ appearing in the expression with a nonzero coefficient is called the degree of the polynomial.

Examples

$4x^3-x^2-5$ \gap[10] has degree 3. Degree 3 polynomials are called cubics. \\ \\ $x^4-1$ \gap[10] has degree 4. Degree 4 polynomials are called cuartics. \\ \\ $4x^5-x$ \gap[10] has degree 5. Degree 5 polynomials are called quintics.

Some for you

A polynomial equation of degree $n$ is an equation that can be written in the form

Degree n polymonial = 0

Examples

$4x^3-x^2-5 = 0$ \gap[10] is a degree 3 polymonial equation. Degree 3 polynomial equations are called cubic equations. \\ \\ $x^4-1 = 0$ \gap[10] is a degree 4 polymonial equation. Degree 4 polynomial equations are called quartic equations. \\ \\ $4x^5-x = 0$ \gap[10] is a degree 5 polymonial equation. Degree 5 polynomial equations are called quintic equations.

Solving cubic equations

By definition, a cubic equation can be written in the form

$a, b, c,$ %and $d$ are fixed numbers and $a \neq 0.$

Unlike the case with quadratic equations, cubic equations always have at least one real solution. However, solving a cubic equation gets us into something of a bind: While there is a perfectly respectable formula for the solutions, it is somewhat complicated and involves the use of complex numbers rather heavily. (A detailed account can be found here.) So we restrict attention to cubic equations at least one of whose solutions is a rational number.

We start with a special case that quickly reduces to solving a quadratic equation:

Solution of ax³ + bx² + cx + d = 0 when d = 0
Eg. x³ − 2x² + x = 0

In this case, $x$ is a common factor:

$ax^3+bx^2+cx = x(ax^2+bx+c)$ \t \gap[40] Eg. $x^3+2x^2+x = x(x^2+2x+1)$

The equation is therefore $x(ax^2+bx+c)= 0$ and its solutions are $x = 0$ and the solutions of the quadratic $ax^2+bx+c = 0,$ if any.

Eg. $x(x^2+2x+1)=0$ \t $\quad \Rightarrow \quad x(x+1)^2 = 0$ \t $\quad \Rightarrow \quad x = 0, \ x = -1$

Example
The cubic equation $3x^3-5x^2+2x = 0$ has $a = 3, b = -5, c = 2, d = 0,$

$2x^3-5x^2+2x = 0 \quad$ \t $\Rightarrow \quad x(2x^2-5x+2) = 0$ \t $\Rightarrow \quad x(2x-1)(x-2) = 0$ \\ \t $\Rightarrow \quad x = 0,\ \ x = \dfrac{1}{2}, \ \ x = 2$

Some for you

General Case:
Solution of ax³ + bx² + cx + d = 0 when at least one solution is rational

1. Try putting $x=\pm\dfrac{\text{Factor of }|d|}{\text{Factor of }|a|}$. \t Eg. $x=\pm\dfrac{\text{Factor of }2}{\text{Factor of }6} = \pm\dfrac{1}{1}$ or $\pm\dfrac{1}{6}$ or $\pm\dfrac{1}{3}$ or $\pm\dfrac{2}{3}$ or $\pm\dfrac{2}{1}$

If there is a rational solution, it will be one of these values; call it $r.$ $(x - r)$ is then a factor of the left-hand side.

2. To get the other factor, you can either divide the given cubic by $(x-r)$ (the other factor will then be the quotient), or use the following formula:

$ax^3+bx^2+cx+d$ \t ${}= (x-r)(ax^2 \ + \ [ar+b]x \ + \ [ar^2+br+c])$

3. Now use factorization or the quadratic formula on the second factor to obtain the remaining solutions (if any).

Example
The cubic equation $3x^3+5x^2-5x+1 = 0$ has $a = 3, b = 5, c = -5, d = 1,$

1. The only factor of $|d| = 1$ is $1$ and the factors of $|a| = 3$ are $1$ and $3$.
Taking all possible ratios $\pm\dfrac{\text{Factor of }|d|}{\text{Factor of }|a|}$ gives

$x=\pm\dfrac{\text{Factor of }1}{\text{Factor of }3} = \pm\dfrac{1}{1}$ or $\pm\dfrac{1}{3}$

$x = 1:$ \t \gap[10] $3(1)^3+5(1)^2-5(1)+1 = 4 \neq 0 $ ✘ \\ $x = -1:$ \t \gap[10] $3(-1)^3+5(-1)^2-5(-1)+1 = 8 \neq 0 $ ✘ \\ $x = \frac{1}{3}:$ \t \gap[10] $3\left(\frac{1}{3}\right)^3+5\left(\frac{1}{3}\right)^2-5\left(\frac{1}{3}\right)+1 = 0 $ ✔

Thus, $x = \frac{1}{3}$ is one solution.

2. $\ \left(x-\frac{1}{3}\right)$ is therefore a factor of the left-hand side $3x^3+5x^2-5x+1.$
To get other solutions (if any), we first factor the left-hand side using the formula above:

$ax^3+bx^2+cx+d$ \t ${}= (x-r)(ax^2 \ + \ [ar+b]x \ + \ [ar^2+br+c])$ \\ $3x^3+5x^2-5x+1$ \t ${}= \left(x-\frac{1}{3}\right)\left(3x^2 \ + \ \left[3\left(\frac{1}{3}\right)+5\right]x \ + \ \left[3\left(\frac{1}{3}\right)^2+5\left(\frac{1}{3}\right)-5\right]\right)$ \\ \t ${}= \left(x-\frac{1}{3}\right)\left(3x^2 \ + \ [1+5]x \ + \ \left[\frac{1}{3}+\frac{5}{3}-5\right]\right)$ \\ \t ${}= \left(x-\frac{1}{3}\right)(3x^2 \ + \ 6x \ - \ 3)$

3. The second factor is

$3x^2 + 6x - 3 = 3(x^2+2x-1).$ Although $x^2+2x-1$ does not factor over the integers, it does have two real solutions given by the quadratic formula:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = \dfrac{-2\pm\sqrt{8}}{2} = -1 \pm \sqrt{2}.$

Thus we have solved the cubic: it has the three real solutions

$x = \dfrac{1}{3},\quad x = -1+\sqrt{2},$ and $x = -1-\sqrt{2}$

Solution of higher-order polynomial equations

Logically speaking, our next step should be a discussion of quartics, then quintics, and so on forever. Well, we've got to stop somewhere, and cubics may be as good a place as any. On the other hand, since we've gotten so far, we ought to at least tell you what is known about higher order polynomials.

Quartics Just as in the case of cubics, there is a formula to find the solutions of quartics. See, for instance, this article.

Quintics and beyond: All good things must come to an end, we're afraid. It turns out that there is no "quintic formula." In other words, there is no single algebraic formula or collection of algebraic formulas that will give the solutions to all quintics. This question was settled by the Norwegian mathematician Niels Henrik Abel in 1824 after almost 300 years of controversy about this question. (In fact, several notable mathematicians had previously claimed to have devised formulas for solving the quintic, but these were all shot down by other mathematicians—this being one of the favorite pastimes of practitioners of our art.) The same negative answer applies to polynomial equations of degree 6 and higher. It's not that these equations don't have solutions, just that they can't be found using algebraic formulas. However, there are certain special classes of polynomial equations that can be solved with algebraic methods. The way of identifying such equations was discovered around 1829 by the French mathematician Évariste Galois.

Now try the exercises in Section 0.5 in Finite Mathematics and Applied Calculus. or move ahead to the next tutorial by pressing "Next tutorial" on the sidebar.