Cubic Equation Formula: An equation is a mathematical statement with an ‘equal to’ sign between two algebraic expressions with equal values. In algebra, there are three types of equations based on the degree of the equation: linear, quadratic, and cubic. We will discuss all these equations and formulas, including the cubic equation formula, in detail here.

A linear equation is one in which the greatest power of the variable or the equation degree is one, while a quadratic equation is one in which the degree of the equation is 2. A cubic equation is one in which the maximum power of the variable or the equation degree is three. Let us learn everything about the formula for the cubic equation in this article.

What is an Equation?

An equation is a mathematical statement that consists of an equal to symbol between two algebraic expressions with the same value.
One or more variables are used in the most fundamental and typical algebraic equations.
For example, \(5e + 8 = 4\) is an equation in which the expressions \(5e + 8\) and \(4\) are separated by an equal sign.
Consider an equation to be a set of weights. You can put different amounts on the left and right sides, but each side must be balanced to answer the problem.
There will always be an unknown in an algebraic equation. A symbol, such as \(x,y\) or \(z,\) is used to indicate this.
Solving an equation involves executing the identical procedure on both sides to determine the value of the unknown integer.
When solving an equation, use the BODMAS acronym to indicate the order of operations: Brackets, Order or Indices, Division, Multiplication, Addition, and Subtraction.

Types of Equations

There are three main types of equations in math.
1. Linear equation
2. Quadratic equation
3. Cubic Equation

Linear Equation

A linear equation is one in which the greatest power of the variable or the equation degree is one. A linear equation’s standard form is \(ax + b,\) where \(a,b\) are constants, and \(a \ne 0.\) The coefficients of \({x^1},{x^0}\) are \(a,b,\) respectively.

Quadratic Equation

A quadratic equation is a two-dimensional equation that may be expressed as \(a{x^2} + bx + c = 0,\) where \(a,b,c \in R\) and \(a \ne 0.\) Every quadratic equation has a corresponding quadratic function, which is obtained by converting the \(”0”\) to a \(”f\left( x \right).”\) A quadratic function is written in standard form as \(f\left( x \right) = a{x^2} + bx + c.\)

Cubic Equation

A cubic equation is one in which the maximum power of the variable or the equation degree is three. The standard form of a cubic equation is \(a{x^3} + b{x^2} + cx + d = 0,\) where \(a,b,c,d\) are constants and \(a \ne 0.\) And \(a,b,c,d\) are the coefficients of \({x^3},{x^2},{x^1},{x^0}\) respectively.

Remainder Theorem

Let \(p\left( x \right)\) be a polynomial of degree one or more and let \(a\) be any real number. If \(p\left( x \right)\) is divided by \(\left({x – a} \right)\) then the remainder is \(p\left( a \right).\)

Factors of a Polynomial

Let \(p\left( x \right)\) be a polynomial of degree one or more and let a be any real number.
(i) If \(p\left( a \right) = 0,\) then \(\left({x – a} \right)\) is a factor of \(p\left( x \right).\)
(ii) If \(\left({x – a} \right)\) is a factor of \(p\left( x \right),\) then \(p\left( a \right) = 0.\)

How to Solve Cubic Equations?

A cubic problem is traditionally solved by reducing it to a quadratic equation and then using factoring or the quadratic formula.

A cubic equation may have three real roots, similar to how a quadratic equation has two. On the other hand, a cubic equation has at least one actual root, unlike a quadratic equation, which may have no real solution at times. The other two roots might be real or imaginary.

When given a cubic equation or any equation, you must always organize it in a standard form first.

If you’re given something like \(5{x^2} + 2x – 5 = \frac{3}{x},\) for example, you’ll re-arrange it into standard form and write it as \(5{x^3} + 2{x^2} – 5x – 3 = 0.\) Then you may solve it using whatever way you choose.
Example: Find the roots of \(f\left( x \right) = {x^3} – 4{x^2} – 6x + 5 = 0.\)
Solution: To find the possible factors, we first find the factors of the constant term, then we put those values and check whether these are satisfying.
Since \(d = 5,\) then the possible factors are \(1\) and \(5.\)
Step 1: First, use the factor theorem to check the possible values by the trial-and-error method.
\(f\left( 1 \right) = 1 – 4 – 6 + 5 \ne 0\)
\(f\left({ – 1} \right) = 1 – 4 + 6 + 5 \ne 0\)
\(f\left( 5 \right) = 125 – 100 – 30 + 5 = 0\)
We find that the root is \(5.\)
Step 2: Find the other roots either by inspection or by the long division method.
\({x^3} – 4{x^2} – 6x + 5 = 0\)
\(\left({x – 5} \right)\left({{x^2} + x – 1}\right) = 0\)
So, the roots are \(x = 5,\frac{{ – 1 + \sqrt 5 }}{2},\frac{{ – 1 – \sqrt 5 }}{2}\)
Example: Check whether \(\left({2x – 3} \right)\) is a factor of \(\left({x + 2{x^3} – 9{x^2} + 12} \right).\)
Let \(p\left( x \right) = 2{x^3} – 9{x^2} + x + 12\) and \(g\left( x \right) = 2x – 3\)
Now, \(g\left( x \right) = 0 \Rightarrow 2x – 3 = 0 \Rightarrow 2x = 3 \Rightarrow x = \frac{3}{2}\)
By factor theorem, \(g\left( x \right)\) will be a factor of \(p\left( x \right),\) if \(p\left({\frac{3}{2}} \right) = 0\)
Now, \(p\left({\frac{3}{2}} \right) = \left\{{2 \times {{\left({\frac{3}{2}} \right)}^3} – 9 \times {{\left({\frac{3}{2}} \right)}^2} + \frac{3}{2} + 12} \right\}\)
\( = \left\{{\left({2 \times \frac{{27}}{8}} \right) – \left({9 \times \frac{9}{4}} \right) + \frac{3}{2} + 12}\right\}\)
\( = \left({\frac{{27}}{4} – \frac{{81}}{4} + \frac{3}{2} + 12} \right) = \left({\frac{{27 – 81 + 6 + 48}}{4}} \right) = \frac{0}{4} = 0\)
Since \(p\left({\frac{3}{2}} \right) = 0,\) so \(g\left( x \right)\) is a factor of \(p\left( x \right).\)

Solving a Cubic Equation Using Graphical Method

We can solve the cubic equation graphically if you can’t solve it using other techniques. We will need an exact drawing of the supplied cubic equation for this.

A solution of the equation is the point(s) where its graph crosses the \(X\)-axis. The number of actual solutions for cubic equations is the number of times their graph crosses the \(X\)-axis.
Example: Find the roots of \({x^3} + 5{x^2} + 2x – 8 = 0\) graphically.
Solution: Simply substitute random values for \(x\) in the graph of the following function:
\(f\left( x \right) ={x^3} + 5{x^2} + 2x – 8\)

\(x\)	\(0\)	\(1\)	\(2\)	\(-2\)
\(f\left( x \right)\)	\( – 8\)	\(0\)	\(24\)	\(0\)

Because the \(x\)-axis of the graph is cut at three locations, there are three actual solutions.

The following are the solutions based on the graph:

\(x = 1,x = \,- 2\,\& \,x = \, – 4.\)

Relation Between Coefficients and Roots of a Cubic Equation

If \(\alpha ,\beta \) and \(\gamma \) are the roots of a cubic equation \(a{x^3} + b{x^2} + cx + d = 0,\) then
The sum of roots \( = \alpha + \beta + \gamma = \, – \frac{{{\text{Coefficient}}\,{\text{of}}\,{x^2}}}{{{\text{Coefficient}}\,{\text{of}}\,{x^3}}} = \,- \frac{b}{a}\)
The sum of the product of roots \( = \alpha \beta + \beta \gamma + \gamma \alpha = \frac{{{\text{Coefficient}}\,{\text{of}}\,x}}{{{\text{Coefficient}}\,{\text{of}}\,{x^3}}} = \frac{c}{a}\)
The product of roots \( = \alpha \beta \gamma = \, – \frac{{{\text{Constant}}\,{\text{term}}}}{{{\text{Coefficient}}\,{\text{of}}\,{x^3}}} = \,- \frac{d}{a}\)

Proof: A cubic equation whose roots are \(\alpha ,\beta ,\) and \(\gamma \) is \(\left({x – \alpha } \right)\left({x – \beta }\right)\left({x – \gamma } \right) = 0.\) Since we can represent it in the form \(a{x^3} + b{x^2} + cx + d = 0 \Rightarrow {x^3} + \left({\frac{b}{a}} \right){x^2} + \left({\frac{c}{a}} \right)x + \left({\frac{d}{a}} \right) = 0,\) The following is our strategy:
\(\left({x – \alpha } \right)\left({x – \beta } \right)\left({x – \gamma } \right) = 0\)
\(\left({{x^2} – \beta x + \alpha x + \alpha \beta } \right)\left({x – \gamma } \right) = 0\)
\( = {x^3}\beta {x^2} – – \alpha {x^2} + \alpha \beta \gamma – \gamma {x^2} + \beta \gamma x + \alpha \gamma x – \alpha \beta \gamma = 0\)
\({x^3} + \left\{{ – \left({\alpha + \beta + \gamma } \right)}\right\}{x^2} + \left({\alpha \beta + \beta \gamma + \gamma \alpha }\right)x + \left({ – \alpha \beta \gamma } \right) = 0\)
On comparing with the equation \({x^3} + \left({\frac{b}{a}} \right){x^2} + \left({\frac{c}{a}} \right)x + \left({\frac{d}{a} = 0,} \right)\) we get
\(\alpha + \beta + \gamma = \,- \frac{b}{a}\)
\(\alpha \beta + \beta \gamma + \gamma \alpha = \frac{{\text{c}}}{{\text{a}}}\)
\(\alpha \beta \gamma = \,- \frac{d}{a}\)

Example: Find a cubic equation with the sum, sum of the product of its roots taken two at a time, and the product of its roots as \(2, – 7, – 14,\) respectively.
Let the equation be \(a{x^3} + b{x^2} + cx + d\) and the roots are \(\alpha ,\beta \) and \(\gamma \)
Then, the sum of roots \(\alpha + \beta + \gamma = 2\)
The sum of the product of roots \(\alpha \beta + \beta \gamma + \gamma \alpha = – 7\)
The product of roots \(\alpha \beta \gamma + = \, – 14\)
We know that the cubic equation is given by
\({x^3} – \left({{\text{sum}}\,{\text{of}}\,{\text{roots}}} \right){x^2} + \left({{\text{the}}\,{\text{sum}}\,{\text{of}}\,{\text{the}}\,{\text{product}}\,{\text{of}}\,{\text{root}}} \right)x – {\text{product}}\,{\text{of}}\,{\text{roots}} = 0\)
\({x^3} = 2{x^2} – 7x + 14 = 0\)
So, one cubic equation which satisfies the given conditions will be \({x^3} – 2{x^2} – 7x + 14 = 0.\)

Solved Examples on Cubic Equation Formula

Q.1. Find the roots of the cubic equation \(3{x^3} – 3{x^2} – 90x = 0\)
Ans: Given, \(3{x^3} – 3{x^2} – 90x = 0\)
Take out \(3x\) as a common factor
\(3x\left({{x^2} – x – 30} \right) = 0\)
\(3x\left({{x^2} – 6x + 5x – 30} \right) = 0\)
\(3x\left({x – 6} \right)\left({x + 5} \right) = 0\)
\( \Rightarrow x = 0,6, – 5\)
Hence, the roots of the given equation are \(0,6, – 5.\)

Q.2. Solve the equation \({x^3} – 6{x^2} + 11x – 6 = 0\) graphically.
Ans: We want to solve the cubic equation \({x^3} – 6{x^2} + 11x – 6 = 0\)
We can factorize this equation to give \(\left({x – 1} \right)\left({x – 2} \right)\left({x – 3} \right) = 0\)
The given cubic equation has \(3\) real and distinct roots.
The solutions of the given equation are \(x = 1,x = 2,\) and \(x = 3.\)

It starts low on the left because \({x^3}\) becomes larger and negative as \(x\) gets larger and negative, and it concludes higher on the right because \({x^3}\) gets larger and positive as \(x\) goes larger and positive. Three times the curve crosses the \(X\)-axis, once at \(x = 1,\) once at \(x = 2,\) and once at \(x = 3.\) As a result, we have three distinct solutions.

Q.3. Solve the cubic equation \({x^3} – 4{x^2} – 9x + 36 = 0\)
Ans: Given, \({x^3} – 4{x^2} – 9x + 36 = 0\)
\({x^2}\left({x – 4} \right) – 9\left({x – 4} \right) = 0\)
Take out the common factor
\(\left({x – 4} \right)\left({{x^2} – 9} \right) = 0\)
\(\left({x – 4} \right)\left({x + 3} \right)\left({x – 3} \right) = 0\)
\(x = 3,4, – 3\)
Hence, the solutions of the given equation are \(x = 3,3,4\)

Q.4. Obtain the roots of the cubic equation \({x^3} – 6{x^2} – 6x – 7 = 0\)
Ans: Given, \({x^3} – 6{x^2} – 6x – 7 = 0\)
Since \(d = 7,\) then the possible factors are \( \pm 1\) and \( \pm 7.\)
Step 1: First, use the factor theorem to check the possible values by the trial-and-error method.
\(f\left( 1 \right) = 1 – 6 – 6 – 7 \ne 0\)
\(f\left({ – 1} \right) = \, – 1 – 6 + 6 – 7 \ne 0\)
\(f\left( 7 \right) = 343 – 294 – 42 – 7 = 0\)
We find that the root is \(7.\)
Step 2: Find the other roots either by inspection or by the long division method. \({x^3} – 6{x^2} – 6x – 7 = 0\)

\(\left({x – 7} \right)\left({{x^2} + x + 1} \right) = 0\)
So, the roots are \(x = 7,\frac{{ – 1 + i\sqrt 3 }}{2},\frac{{ – 1 – i\sqrt 3 }}{2}\)

Q.5. Samaira wanted to solve this equation \({x^3} – 3{x^2} – 3x + 1 = 0\) and find the roots.
Ans: Given equation is \({x^3} – 3{x^2} – 3x + 1 = 0\)
We know that, \({\left({a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}\)
So, on comparing the given equation with this identity, we get
\(a = x\) and \(b = 1\)
Therefore, \({\left({x + 1} \right)^3} = 0\)
\(x = \,- 1, – 1, – 1\)
Hence, the given equation has three equal roots \( – 1, – 1, – 1.\)

Summary on Solving a Cubic Equation

In this article, we looked at what a cubic equation is. A cubic equation is an algebraic equation with the degree of equation \(3.\) Then we went over how to solve a cubic equation. We also spoke about how to use a graphical strategy for solving a cubic equation problem. Finally, we uncovered the relationship between the coefficients and roots of a cubic equation, as well as its proof. Solved examples can help you grasp the cubic equation formula better.

FAQs on Formula for Cubic Equation

Here are some most frequently asked questions on cubic equation formula:

Q.1: What is a cubic equation formula?
Ans: A cubic equation is an algebraic equation of degree three. The standard form of a cubic equation is defined as \(a{x^3} + b{x^2} + cx + d = 0,\) where \(a,b,c,d\) are integers and \(a\) is non-zero. Cubic equations always have three roots, some of which may be equal, according to the fundamental theorem of algebra.

Q.2: How can we find the roots of the cubic equation?
Ans: The following are the ways to find the roots of the cubic equation:
1. Finding Integer Solutions with Factor Lists
2. Using a Graphical Approach

Q.3: What is the formula to find cubic polynomial?
Ans: A cubic equation whose roots are \(\alpha ,\beta ,\) and \(\gamma \) is
\(\left({x – \alpha } \right)\left({x – \beta } \right)\left({x – \gamma } \right) = 0\)
\(\left({{x^2} – \beta x – \alpha x + \alpha \beta } \right)\left({x – \gamma } \right) = 0\)
\({x^3} – \beta{x^2} – – \alpha {x^2} + \alpha \beta x – \gamma {x^2} + \beta \gamma x + \alpha \gamma x – \alpha \beta \gamma = 0\)
\({x^3} – \left({\alpha + \beta + \gamma } \right){x^2} + \left({\alpha \beta + \beta \gamma + \gamma \alpha } \right)x – \alpha \beta \gamma = 0\)
Hence, the cubic polynomial can be found using the formula
\({x^3} – \left({{\text{sum}}\,{\text{of}}\,{\text{roots}}} \right){x^2} + \left({{\text{the}}\,{\text{sum}}\,{\text{of}}\,{\text{the}}\,{\text{product}}\,{\text{of}}\,{\text{root}}} \right)x – {\text{product}}\,{\text{of}}\,{\text{roots}} = 0\).

Q.4: How many roots are there in a cubic equation?
Ans: There are three roots in a cubic equation. The following cases are possible for the roots of a cubic equation:
1. All three roots might be real and distinct.
2. All three roots might be real, and two of them might be equal.
3. All three roots might be real and equal.
4. One root might be real, and the other two are non-real (complex).

Q.5: How do you know if an equation is a cubic?
Ans: If the maximum degree of the equation is three, then it is a cubic equation.

We hope this detailed article on the cubic equation formula helped you in your studies. If you have any doubts or queries on the cubic equation roots formula or have suggestions regarding this article, feel to ask us in the comment section, and we will be more than happy to assist you. Happy learning!