# How to Solve Cubic and Quartic Equations

Earlier, you learned to solve linear equations (equations where the highest power is 1) and quadratic equations (equations where the highest power is 2). In this section you’ll learn to solve equations with powers of all possible values. You’ll mainly look at cubic and quartic equations—the method is the same for both.

Theory

### TypesofCubicEquations

1.
The equation has $x$ in every term:
 $a{x}^{3}+b{x}^{2}+cx=0.$
2.
The equation has a constant:
 $a{x}^{3}+b{x}^{2}+cx+d=0.$

Rule

### SolvingCubicEquationswith$x$inEveryTerm

1.
Factorize $x$ outside the brackets, $x\phantom{\rule{-0.17em}{0ex}}\left(a{x}^{2}+bx+c\right)=0$.
2.
Factorize the quadratic equation, $\phantom{\rule{-0.17em}{0ex}}\left(x-{x}_{1}\right)\phantom{\rule{-0.17em}{0ex}}\left(x-{x}_{2}\right)=0$.
3.
The factorization of the cubic equation is then
 $x\phantom{\rule{-0.17em}{0ex}}\left(x-{x}_{1}\right)\phantom{\rule{-0.17em}{0ex}}\left(x-{x}_{2}\right)=0.$
4.
Solve the equation with the zero product property ($a\cdot b=0$ if $a=0$ or $b=0$). You then get that $x=a$, $x={x}_{1}$ and $x={x}_{2}$.

Example 1

Solve the equation ${x}^{3}+2{x}^{2}+x=0$

1.
You’ll follow the method above and factorize $x$ outside the brackets: $\begin{array}{llll}\hfill {x}^{3}+2{x}^{2}+x& =0,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}+2x+1\right)& =0.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
2.
Now, factorize the quadratic equation with either the quadratic formula, or with inspection. $\begin{array}{llll}\hfill {x}^{2}+2x+1& =0,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \phantom{\rule{-0.17em}{0ex}}\left(x-1\right)\phantom{\rule{-0.17em}{0ex}}\left(x-1\right)& =0.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
3.
Factorization of the cubic equation is then $\begin{array}{lll}\hfill x\phantom{\rule{-0.17em}{0ex}}\left(x-1\right)\phantom{\rule{-0.17em}{0ex}}\left(x-1\right)=0.& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$
4.
Then you solve the equation:

Example 2

Solve the equation ${x}^{4}=9{x}^{2}$

$\begin{array}{llll}\hfill {x}^{4}& =9{x}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{4}-9{x}^{2}& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}-9\right)& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Now you use the zero product property to find the solutions:

Rule

### SolvingCubicEquationswithaConstant

1.
Guess a solution, and use polynomial long division.
2.
Factorize the cubic equation, $\phantom{\rule{-0.17em}{0ex}}\left(x-a\right)\phantom{\rule{-0.17em}{0ex}}\left(a{x}^{2}+bx+c\right)=0$.
3.
Factorize the quadratic equation, $\phantom{\rule{-0.17em}{0ex}}\left(x-{x}_{1}\right)\phantom{\rule{-0.17em}{0ex}}\left(x-{x}_{2}\right)$.
4.
The factorization of the cubic equation is then
 $\phantom{\rule{-0.17em}{0ex}}\left(x-a\right)\phantom{\rule{-0.17em}{0ex}}\left(x-{x}_{1}\right)\phantom{\rule{-0.17em}{0ex}}\left(x-{x}_{2}\right)=0.$
5.
Solve the equation with the zero product property ($a\cdot b=0$ if $a=0$ or $b=0$). You’ll then get $x=0$, $x={x}_{1}$ and $x={x}_{2}$.

Example 3

Solve the equation ${x}^{3}-7x=-6$

1.
First you move all the terms over to the left side. $\begin{array}{llll}\hfill {x}^{3}-7x& =-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{3}-7x+6& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Now, you’ll have to guess a solution. Begin with $x=1$:

 $\phantom{\rule{-0.17em}{0ex}}{\left(1\right)}^{3}-7\phantom{\rule{-0.17em}{0ex}}\left(1\right)+6=1-7+6=0.$

Lucky for you, the first solution you guessed was correct. (You often start with 1 when you guess a solution, and this is why).

Now you have to use polynomial long division on the equation with $\phantom{\rule{-0.17em}{0ex}}\left(x-1\right)$. Because the expression lacks the ${x}^{2}$-term, you put in an extra space where the ${x}^{2}$-term would have been, or put 0 in front of ${x}^{2}$ in the long polynomial division. It’s easier to maintain control over the terms this way.

2.
The factorization is then
 $\phantom{\rule{-0.17em}{0ex}}\left(x-1\right)\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}+x-6\right).$
3.
You now have to factorize the quadratic expression ${x}^{2}+x-6$ either with the quadratic formula, or with inspection. When you solve the equation
 ${x}^{2}+x-6=0$

you get the answers $x=2$ and $x=-3$. Put them in the factorization formula $a\phantom{\rule{-0.17em}{0ex}}\left(x-{x}_{1}\right)\phantom{\rule{-0.17em}{0ex}}\left(x-{x}_{2}\right)$ so that the factorization becomes

 $\phantom{\rule{-0.17em}{0ex}}\left(x-2\right)\phantom{\rule{-0.17em}{0ex}}\left(x+3\right).$
4.
Now you can find the factorization of the cubic equation by making a product of the factors:
$\begin{array}{llll}\hfill {x}^{3}-7x+6& =\phantom{\rule{-0.17em}{0ex}}\left(x-1\right)\phantom{\rule{-0.17em}{0ex}}\left(x-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\cdot \phantom{\rule{-0.17em}{0ex}}\left(x+3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{lll}\hfill {x}^{3}-7x+6=\phantom{\rule{-0.17em}{0ex}}\left(x-1\right)\phantom{\rule{-0.17em}{0ex}}\left(x-2\right)\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

5.
Then you just have to solve the equation: