Stuck in finding the roots of cubic equations? Do not worry because we are going to learn how to solve cubic equations easily.

Cubic equations might seem a bit tricky and difficult when you look at it and try to solve it the first time. But, if we know the proper method and the basics like how to solve quadratic equations and factorization, you will be able to solve them pretty easily. Today, we are going to learn the method to solve them. You will see at the end of reading this tutorial that if learnt right, they are not difficult at all.

## What is a cubic equation

Before finding the roots of a cubic equation it is important to know what is meaning of cubic equation. A cubic equation is a third-degree polynomial and the standard form of the cubic equation is given as:

**ax ^{3} + bx^{2} +cx + d = 0**

Here, a, b, c and are all real numbers and a cannot be zero because if a is zero, the polynomial will no longer have 3 as the highest power.

For example,

- 5x
^{3}+ 3x^{2}+2x – 1 = 0 is a cubic polynomial with a = 5, b = 3, c = 2 and d = -1 - 4x
^{3}– 5x + 4 = 0 is a cubic equation with a = 4, b = 0, c = -5 and d = 4 - 3x
^{2}+4x + 5 = 0 is not a cubic polynomial because the highest power is not 3. It is a 2^{nd}degree polynomial.

**Number of Roots in a cubic equation and their nature**

Total number of roots in a polynomial is equal to the highest degree present in that polynomial. Therefore, there will be 3 roots for the cubic equation as the highest degree or the power is 3. The root is the value at which the graph touches the x-axis. It is equivalent to finding the solution of a cubic equation.

A cubic polynomial has either one real or three real roots. However, they can be repeated. But, they do not have all the roots imaginary unlike, quadratic equation i.e. the graph of the cubic polynomial will pass through the x-axis at least once.

**Roots of a cubic equation**

In this section you will know how to find the roots of a cubic equation from various methods.

Consider the standard form of cubic equation

ax^{3} + bx^{2} +cx + d = 0

Let us first consider the simplest case, in which the value of the constant d = 0.

ax^{3} + bx^{2} +cx = 0

Let us understand it by solving a sample example.

**1) Find the solution to the cubic polynomial given below:**

**3x ^{3} + 6x^{2} – 9x = 0**

We can take x common from the above equation.

x (3x^{2} + 6x – 9) = 0

We can solve this question easily because 3x^{2} + 6x – 9 is a quadratic equation and we know how to solve it.

Factoring 3x^{2} + 6x – 9

x [3x^{2} + 9x – 3x – 9] = 0

x [3x (x + 3) – 3 (x + 3)] = 0

x (x + 3) (3x – 3) = 0

We have now

x = 0, x + 3 = 0 and 3x – 3 =0

x = 0, x = -3 and x = 1

**Therefore, the roots of the above cubic equation are 0, -3 and 1.**

Above example is the simplest because the constant term is zero. But the real question is how to find the roots of a cubic equation if constant is not equal to zero.

Let us now consider the case where d is not zero.

Consider the standard form of equation again,

ax^{3} + bx^{2} +cx + d = 0

**Steps to find the roots of a cubic equation**

The steps given below explains the method to obtain the roots of the cubic polynomial

- First, find the factors of a and d.
- Create a set A whose members are obtained by dividing the factors of a by each of the factors of d.
- Also, add the negative of the members in the set A.

The integer solution of the cubic polynomial will be from this set.

Two approaches exist to find the remaining roots.

**Trial and error method to solve cubic equations**

This is a tedious and complex approach because what it requires is to guess the values and substitute in the equation to check whether it satisfies the equation or not. It does not require any special skill or method to learn but using this approach may take quite a lot of time to find the solution. This method is generally used to find one of the root of cubic equation and remaining two can be found by synthetic division method.

**Factor Theorem and Synthetic Division Method **

Factor theorem is a better approach to find the roots of a cubic equation as it is a systematic way to find the solution. It might seem difficult at first but once you learn it you can save yourself a lot of trouble.

In this method, the first root is found by trial and error approach taking integer values from the set A. After finding the first root we apply the **factor theorem** which says that if x = p is the root of the equation the (x – p) is the factor of the equation.

Consider the standard form of cubic equation

ax^{3} + bx^{2} +cx + d = 0

(x – p) is the factor of the equation. Then the equation becomes:

**(x – p)(ax ^{2} + bx + c) = 0**

The value of a, b, and c can be found by the synthetic division method.

Let us consider an example to understand this method

#### Find the roots of the following cubic polynomial:

**2x ^{3} + 2x^{2} – 4x – 4 = 0**

**Solution**

In this cubic equation a = 2 and d = -4

- Factors of a: 1 and 2
- Factors of d: 1, 2, and 4
- Dividing the factors of a by the factors of d. Thus, A = {1, ½, ¼, 2}.
- Adding the negatives of the members to the set i.e. A = {1, -1, ½, -1/2, ¼,-1/4, 2, -2}

The integer root will be from this set and can be found by trial and error approach. Plugging integer values from this set into the equation to find the solution.

Let us first guess 1.

2x^{3} + 2x^{2} – 4x – 4 =

2(1)^{3} + 2(1)^{2} – 4(1) – 4 =

2 + 2 – 4 – 4 = -4

-4 ≠ 0

Since the equation is not satisfied, 1 is not the root.

Trying -1

2(-1)^{3} + 2(-1)^{2} – 4(-1) – 4 = 0

-2 + 2 + 4 – 4 = 0

Since the equation is satisfied, x = -1 is the root of the above given question.

**Factorization **

Now, from factor theorem, (x + 1) is the factor of the equation. We have now,

(x + 1)(ax^{2} + bx + c) = 0

**Synthetic Division**

We have to find the value of a, b, and c to obtain the other roots. For this purpose, the synthetic division method is used.

The coefficients of the given equation are a = 2, b = 2, c = -4 and d = -4

Write them as shown in the table below:

2 | 2 | -4 | -4 | x = -1 |

2 |

Take the first number and write it in the 3^{rd} line or row as shown above.

Now, multiply x i.e. -1 by the term in the third row i.e. 3 and write it in 2^{nd} row and 2^{nd} column (below the term b).

Add both terms and write it below as shown below:

2 | 2 | -4 | -4 | x = -1 |

-2 | ||||

2 | 0 |

Again follow the same procedure but this time multiply -1 by 0. We have,

2 | 2 | -4 | -4 | x = -1 |

-2 | 0 | |||

2 | 0 | -4 |

Similarly,

2 | 2 | -4 | -4 | x = -1 |

-2 | 0 | 4 | ||

2 | 0 | -4 | 0 |

If the results add up to zero for the 4^{th} column, then it verifies that x = -1 is the root of the equation. Otherwise, you have done a mistake in finding the value of x.

The value of a, b and c are 2, 0, -4 respectively. Substituting them in the equation, we get

(x + 1) (2x^{2} + 0x – 4) = 0

or

(x + 1) (2x^{2} – 4) = 0

2x^{2} – 4 is a quadratic equation and we can solve it using factoring, completing the square metthod or the quadratic formula.

In this case, we will use factoring method as it is easier. You can use any of them.

2x^{2} – 4 = 0

2x^{2} = 4

x^{2} = 2

x = ± 1.414

**Hence, the solution or roots of the quadratic equation are x = -1, 1.414, -1.414.**

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