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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

Types of Cubic Equations

1.
The equation has x in every term:
ax3 + bx2 + cx = 0.
2.
The equation has a constant:
ax3 + bx2 + cx + d = 0.

Rule

Solving Cubic Equations with x in Every Term

1.
Factorize x outside the brackets, x (ax2 + bx + c) = 0.
2.
Factorize the quadratic equation, (x x1) (x x2) = 0.
3.
The factorization of the cubic equation is then
x (x x1) (x x2) = 0.
4.
Solve the equation with the zero product property (a b = 0 if a = 0 or b = 0). You then get that x = a, x = x1 and x = x2.

Example 1

Solve the equation x3 + 2x2 + x = 0

1.
You’ll follow the method above and factorize x outside the brackets: x3 + 2x2 + x = 0, x (x2 + 2x + 1) = 0.
2.
Now, factorize the quadratic equation with either the quadratic formula, or with inspection. x2 + 2x + 1 = 0, (x 1) (x 1) = 0.
3.
Factorization of the cubic equation is then x (x 1) (x 1) = 0.
4.
Then you solve the equation: x (x 1) (x 1) = 0 x = 0 and x = 1

Example 2

Solve the equation x4 = 9x2

x4 = 9x2 x4 9x2 = 0 x2 (x2 9) = 0 x2 (x 3) (x + 3) = 0 Now you use the zero product property to find the solutions:

x2 (x 3) (x + 3) = 0 x2 = 0 x = 0 and x = 3 and x = 3

x2 (x 3) (x + 3) = 0 x2 = 0 x = 0 and x = 3 and x = 3

Rule

Solving Cubic Equations with a Constant

1.
Guess a solution, and use polynomial long division.
2.
Factorize the cubic equation, (x a) (ax2 + bx + c) = 0.
3.
Factorize the quadratic equation, (x x1) (x x2).
4.
The factorization of the cubic equation is then
(x a) (x x1) (x x2) = 0.
5.
Solve the equation with the zero product property (a b = 0 if a = 0 or b = 0). You’ll then get x = 0, x = x1 and x = x2.

Example 3

Solve the equation x3 7x = 6

1.
First you move all the terms over to the left side. x3 7x = 6 x3 7x + 6 = 0

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

(1) 3 7 (1) + 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 (x 1). Because the expression lacks the x2-term, you put in an extra space where the x2-term would have been, or put 0 in front of x2 in the long polynomial division. It’s easier to maintain control over the terms this way.

Polynomial long division of x^3-7x+6 divided by x-1

2.
The factorization is then
(x 1) (x2 + x 6) .
3.
You now have to factorize the quadratic expression x2 + x 6 either with the quadratic formula, or with inspection. When you solve the equation
x2 + x 6 = 0

you get the answers x = 2 and x = 3. Put them in the factorization formula a (x x1) (x x2) so that the factorization becomes

(x 2) (x + 3) .
4.
Now you can find the factorization of the cubic equation by making a product of the factors:
x3 7x + 6 = (x 1) (x 2) (x + 3)

x3 7x + 6 = (x 1) (x 2) (x + 3)

5.
Then you just have to solve the equation: (x 1) (x 2) (x + 3) = 0 x = 1 and x = 2 and x = 3

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What Is Polynomial Long Division?