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Solving Cubic Equations – Methods & Examples

Solving higher order polynomial equations is an essential skill for anybody studying science and mathematics. However, understanding how to solve these kind of equations is quite challenging.

In this article, we are going to learn how solve the cubic equations using different methods such as the division method, Factor Theorem and factoring by grouping.

But before getting into this topic, let’s discuss what a polynomial and cubic equation is.

A polynomial is an algebraic expression with one or more terms in which a constant and a variable are separated by an addition or a subtraction sign.

The general form of a polynomial is axn + bxn-1 + cxn-2 + …. + kx + l, where each variable has a constant accompanying it as its coefficient. The different types of polynomials include; binomials, trinomials and quadrinomial. Examples of polynomials are; 3x + 1, x2 + 5xy – ax – 2ay, 6x2 + 3x + 2x + 1 etc.

A cubic equation is an algebraic equation of third-degree.
The general form of a cubic function is: f (x) = ax3 + bx2 + cx1 + d. And the cubic equation has the form of ax3 + bx2 + cx + d = 0, where a, b and c are the coefficients and d is the constant.

How to Solve Cubic Equations?

The traditional way of solving a cubic equation is to reduce it to a quadratic equation and then solve either by factoring or quadratic formula.

Like a quadratic equation has two real roots, a cubic equation may have possibly three real roots. But unlike quadratic equation which may have no real solution, a cubic equation has at least one real root.

The other two roots might be real or imaginary.

Whenever you are given a cubic equation, or any equation, you always have to arrange it in a standard form first.

For example, if you are given something like this, 3x2 + x – 3 = 2/x, you will re-arrange into the standard form and write it like, 3x3 + x2 – 3x – 2 = 0. Then you can solve this by any suitable method.

Let’s see a few examples below for better understanding:

Example 1

Determine the roots of the cubic equation 2x3 + 3x2 – 11x – 6 = 0

Solution

Since d = 6, then the possible factors are 1, 2, 3 and 6.

Now apply the Factor Theorem to check the possible values by trial and error.

f (1) = 2 + 3 – 11 – 6 ≠ 0
f (–1) = –2 + 3 + 11 – 6 ≠ 0
f (2) = 16 + 12 – 22 – 6 = 0

Hence, x = 2 is the first root.

We can get the other roots of the equation using synthetic division method.
= (x – 2) (ax2 + bx + c)
= (x – 2) (2x2 + bx + 3)
= (x – 2) (2x2 + 7x + 3)
= (x – 2) (2x + 1) (x +3)

Therefore, the solutions are x = 2, x = -1/2 and x = -3.

Example 2

Find the roots of the cubic equation x3 − 6x2 + 11x – 6 = 0

Solution

x3 − 6x2 + 11x – 6

(x – 1) is one of the factors.

By dividing x3 − 6x2 + 11x – 6 by (x – 1),

⟹ (x – 1) (x2 – 5x + 6) = 0

⟹ (x – 1) (x – 2) (x – 3) = 0

This of the cubic equation solutions are x = 1, x = 2 and x = 3.

Example 3

Solve x3 – 2x2 – x + 2

Solution

Factorize the equation.

x3 – 2x2 – x + 2 = x2(x – 2) – (x – 2)

= (x2 – 1) (x – 2)

= (x + 1) (x – 1) (x – 2)

x = 1, -1 and 2.

Example 4

Solve the cubic equation x3 – 23x2 + 142x – 120

Solution

First factorize the polynomial.

x3 – 23x2 + 142x – 120 = (x – 1) (x2 – 22x + 120)

But x2 – 22x + 120 = x2 – 12x – 10x + 120

= x (x – 12) – 10(x – 12)
= (x – 12) (x – 10)

Therefore, x3 – 23x2 + 142x – 120 = (x – 1) (x – 10) (x – 12)

Equate each factor to zero.

x – 1= 0

x = 1

x – 10 = 10

x – 12= 0

x = 12

The roots of the equation are x = 1, 10 and 12.

Example 5

Solve the cubic equation x3 – 6 x2 + 11x – 6 = 0.

Solution

To solve this problem using division method, take any factor of the constant 6;

let x = 2

Divide the polynomial by x-2 to

(x2 – 4x + 3) = 0.

Now solve the quadratic equation (x2 – 4x + 3) = 0 to get x= 1 or x = 3

Therefore, the solutions are x = 2, x= 1 and x =3.

Example 6

Solve the cubic equation x3 – 7x2 + 4x + 12 = 0

Solution

Let f(x) = x3 – 7x2 + 4x + 12

Since d = 12, the possible values are 1, 2, 3, 4, 6 and 12.

By trial and error, we find that f (–1) = –1 – 7 – 4 + 12 = 0

So, (x + 1) is a factor of the function.

x3 – 7x2 + 4x + 12
= (x + 1) (x2 – 8x + 12)
= (x + 1) (x – 2) (x – 6)

Therefore x = –1, 2, 6

Example 7

Solve the following cubic equation:

x3 + 3x2 + x + 3 = 0.

Solution

x3 + 3x2 + x + 3
= (x3 + 3x2) + (x + 3)
= x2(x + 3) + 1(x + 3)
= (x + 3) (x2 + 1)

Therefore, x = -1 ,1 -3.

Example 8

Solve x3 − 6x2 + 11x − 6 = 0

Solution

Factorize

x3 − 6x2 + 11x − 6 = 0 ⟹ (x − 1) (x − 2) (x − 3) = 0

Equating each factor to zero gives;

x = 1, x = 2 and x = 3

Example 9

Solve x 3 − 4x2 − 9x + 36 = 0

Solution

Factorize each set of two terms.

x2(x − 4) − 9(x − 4) = 0

Extract the common factor (x − 4) to give

(x2 − 9) (x − 4) = 0

Now factorize the difference of two squares

(x + 3) (x − 3) (x − 4) = 0

By equating each factor to zero, we get;

x = −3, 3 or 4

Example 10

Solve the equation 3x3 −16x2 + 23x − 6 = 0

Solution

Divide 3x3 −16x2 + 23x – 6 by x -2 to get 3x2 – 1x – 9x + 3

= x (3x – 1) – 3(3x – 1)

= (x – 3) (3x – 1)

Therefore, 3x3 −16x2 + 23x − 6 = (x- 2) (x – 3) (3x – 1)

Equate each factor to zero to get,

x = 2, 3 and 1/3

Example 11

Find the roots of 3x3 – 3x2 – 90x=0

Solution

factor it out 3x

3x3 – 3x2 – 90x ⟹3x (x2 – x – 30)

Find a pair of factors whose product is −30 and sum is −1.

⟹- 6 * 5 =-30

⟹ −6 + 5 = -1

Rewrite the equation by replacing the term “bx” with the chosen factors.

⟹ 3x [(x2 – 6x) + (5x – 30)]

Factor the equation;

⟹ 3x [(x (x – 6) + 5(x – 6)]

= 3x (x – 6) (x + 5)

By equating each factor to zero, we get;

x = 0, 6, -5

Solving cubic equations using graphical method

If you are unable to solve the cubic equation by any of the above methods, you can solve it graphically. For that, you need to have an accurate sketch of the given cubic equation.

The point(s) where its graph crosses the x-axis, is a solution of the equation. The number of real solutions of the cubic equations are same as the number of times its graph crosses the x-axis.

Example 12

Find the roots of x3 + 5x2 + 2x – 8 = 0 graphically.

Solution

Simply draw the graph of the following function by substituting random values of x:

f (x) = x3 + 5x2 + 2x – 8

You can see the graph cuts the x-axis at 3 points, therefore, there are 3 real solutions.

From the graph, the solutions are:

x = 1, x = -2 & x = -4.

Practice Questions

Solve the following cubic equations:

  1. x3 − 4x2 − 6x + 5 = 0
  2. 2x3 − 3x2 − 4x − 35 = 0
  3. x3 − 3x2 − x + 1 = 0
  4. x3 + 3x2 − 6x − 8 = 0
  5. x3 + 4x2 + 7x + 6 = 0
  6. 2x3 + 9x2 + 3x − 4 = 0
  7. x3 + 9x2 + 26x + 24 = 0
  8. x3 − 6x2 − 6x − 7 = 0
  9. x3 − 7x − 6 = 0
  10. x3 − 5x2 − 2x + 24 =0
  11. 2x3 + 3x2 + 8x + 12 = 0
  12. 5x3 − 2x2 + 5x − 2 = 0
  13. 4x3 + x2 − 4x − 1 = 0
  14. 5x3 − 2x2 + 5x − 2 = 0
  15. 4x3− 3x2 + 20x − 15 = 0
  16. 3x3 + 2x2 − 12x − 8 = 0
  17. x3 + 8 = 0
  18. 2x3 − x2 + 2x − 1 = 0
  19. 3x3 − 6x2 + 2x − 4 = 0
  20. 3x3 + 5x2 − 3x − 5 = 0

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