Cubic Equation is a mathematical equation in which a polynomial of degree 3 is equated to a constant or another polynomial of maximum degree 2. The standard representation of the cubic equation is ax3+bx2+cx+d = 0 where a, b, c, and d are real numbers. Some examples of cubic equation are x3 - 4x2 + 15x - 9 = 0, 2x3 - 4x2 = 0 etc.
For learning How to Solve Cubic Equations we must first learn about polynomials, the degree of the polynomial, and others. In this article, we will learn about, Polynomials, Polynomial Equations, Solving Cubic Equations Or how to solve cubic equations, and others in detail.
Polynomial Definition
Polynomial is defined as follows,
A polynomial is an algebraic expression in which the power of a variable is a non-negative integer. The general form of a polynomial is a0xn + a1xn-1 + a2xn-2 +... + an. Depending upon the maximum power of the variable, a polynomial can be classified as a monomial, binomial, trinomial, and so on.
What is an Equation?
An Equation is defined as follows,
An equation is a polynomial that is equated to a numerical value or any other polynomial. For Example, x + 2 is a polynomial but x + 2 = 5 is an equation. Similarly, 2x + 3 = x + 1 is also an equation whereas, 2x + 3 and x + 1 are polynomials individually.
Degree of Equation
The definition of the Degree of Equation is stated below:
Degree of a Equation is defined as the maximum power possessed by the variable in an Equation.
Based on the degree of the Equation, an Equation can be classified as follows:
- Linear Equation
- Quadratic Equation
- Cubic Equation
- Biquadratic Equation
Linear Equation
The Equation in which the maximum power of the variable is 1 is called a Linear Equation.
For example: 3x + 1 = 0
Quadratic Polynomial
The Equation in which the maximum power of the variable is 2 is a Quadratic Equation.
For example: 3x2 + x + 1 = 0
Cubic Equation
The Equation in which the maximum power of the variable is 3 is called a Cubic Equation.
For example: 5x3 + 3x2 + x + 1 = 0
Biquadratic Polynomial
The Equation in which the maximum power of the variable is 4 is called a Biquadratic Polynomial or Quartic Polynomial.
For example: 5x4+4x3+3x2+2x+1 = 0
Cubic Equation Definition
Cubic Equation is an algebraic equation where the highest degree of the polynomial is 3.
Some examples of cubic equations are 5x3 + 3x2 + x + 1 = 0, 2x3 + 8 = x ⇒ 2x3 - x + 8 = 0, etc.
The general form of a cubic equation is,
ax3 + bx2 + cx + d = 0, a ≠ 0
Where,
- a, b, and c are the coefficients of variable and their exponenats and d is the constant, and
- a, b, c and d are real numbers.
How to Solve Cubic Equations?
A cubic equation is a equation with degree three. It has three solutions and it can be solved easily by following the steps added below,
Step 1: Find one solution to the cubic equation by hit and try method. Suppose we have a cubic equation P(x) then find for any x = a, P(a) = 0 by taking, x = 0, ±1, ±2, ±3, ... and so.
Step 2: When we get, P(a) = 0, find the factor (x - a) of P(x)
Step 3: Divide P(x) by (x - a) to get a quadratic equation say Q(x) using polynomial division.
Step 4: Factarize the quadratic equation Q(x) to get the factors as (x - b), and (x - c).
Step 5: (x - a), (x - b), and (x - c) are the factors of P(x) and solving each factors we gets the roots of equation as, a, b, and c.
Learn more about, Dividing Polynomial
Solving Cubic Equations
A Cubic Equation can be solved by two methods
- By reducing it into a quadratic equation and then solving it either by factoring or the quadratic formula
- By Graphical Method
A Cubic Equation has three roots. These roots may be real or imaginary. Also, there can be distinct roots or two same and one different root and all three same roots.
Point to be noted that for any equation, including Cubic Equations, the equation must always be arranged in its standard form first before solving the equation.
For instance, if the given equation is 2x2-5 = x + 4/x, then we have to re-arrange this into its standard form, i.e., 2x3-x2-5x-4 = 0. Now, we can solve the equation using any appropriate method.
Solving Cubic Equation Using Factors
The solution of the cubic equation using the factor theorem is explained using the example added below,
Example: Find the roots of equation f(x) = 3x3 −16x2 + 23x − 6 = 0.
Solution:
Given expression: f(x) = 3x3 −16x2 + 23x − 6 = 0
First, factorize the polynomial to get roots
Since the constant is -6 the possible factors are 1, 2, 3, 6
f(1) = 3 - 16 + 23 - 6 ≠ 0
f(2) = 24 - 64 + 46 - 6 = 0
f(3) = 81 - 144 + 69 - 6 = 0
f(6) = 648 - 576 + 138 - 6 ≠ 0
We know that, according to Factor Theorem if f(a) = 0, then (x-a) is a factor of f(x)
So, (x - 2) and (x - 3) are factors of f(x). Therefore, the product of (x - 2) and (x - 3) will also be factor of f(x). Now to find the remaining factors use the long division method and divide f(x) by product of (x - 2) and (x - 3)
Hence, Divisor = (x - 2)(x - 3) = (x2 - 5x + 6) and Dividend = 3x3 −16x2 + 23x − 6. Now divide as shown below,
After division we obtain (3x- 1) as quotient and remainder is 0. Now as per Division Algorithm we know that
Dividend = Divisor × Quotient + Remainder.
⇒ f(x) = (3x3 −16x2 + 23x − 6) = (x2 - 5x + 6)(3x-1)
Since f(x) = 0
⇒ (x2 - 5x + 6)(3x-1) = 0
⇒ x2 - 5x + 6 = 0 or 3x-1 = 0
Now we will take 3x-1 = 0 ⇒ x = 1/3 as we already know two roots from x2 - 5x + 6 which are 2 and 3
So,Roots of the given Cubic Equation are 1/3, 2, and 3.
Solving Cubic Equation Using Graphical Method
A cubic equation is solved graphically when you cannot solve the given equation using other techniques. So, we need an accurate drawing of the given cubic equation. The equation's roots are the point(s) at which the graph crosses the X-axis if the equation is in the terms of x and if the equation is in the terms of y then the roots of the equation are the points at which the graph cuts the Y-axis.
The number of real solutions to the cubic equation is equal to the number of times the graph of the cubic equation crosses the X-axis.
Example: Find the roots of equation f(x) = x3 − 4x2 − 9x + 36 = 0, using the graphical method.
Solution:
Given expression: f(x) = x3 − 4x2 − 9x + 36 = 0.
Now, simply substitute random values for x in the graph for the given function:
x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
|---|
f(x) | -56 | 0 | 19 | 40 | 36 | 24 | 10 | 0 | 0 | 16 |
|---|
We can see that the graph has cut the X-axis at 3 points, therefore, there are 3 real solutions.
From the graph, the solutions are: x = -3, x = 3, and x = 4.
Hence, the roots of the given equation are -3, 3, and 4.
Read More,
Problems Based on Solving Cubic Equations
Problem 1: Find the roots of f(x) = x3 - 4x2 -3x + 6 = 0.
Solution:
Given expression: f(x) = x3 - 4x2 -3x + 6 = 0.
First, factorize the polynomial to get roots.
Since the constant is +6 the possible factors are 1, 2, 3, 6.
f(1) = 1 - 4 - 3 + 6 = 7 - 7 = 0
f(2) = 8 - 16 - 6 + 6 ≠ 0
f(3) = 27 - 36 - 9 + 6 ≠ 0
f(6) = 216 - 144 -18 + 6 = -48 ≠ 0
So, according to Factor Theorem (x - 1) is a factor of the given equation. Now to find the remaining factors use the long division method.
According to Division Algorithm we can write,
So, f(x) = x3 - 4x2 -3x + 6 = (x - 1) (x2 - 3x - 6) = 0
⇒ (x - 1) = 0 or (x2 - 3x - 6) = 0
We know that the roots of a quadratic equation ax2 + bx + c = 0 are,
x = [-b ± √(b2-4ac)]/2a
Hence, for (x2 - 3x - 6) = 0
x = [3 ± √(32 - 4(1)(-6)]/2(1)
x = (3 ± √33)/2
Hence, the roots of the given cubic equation are 1, (3+√33)/2, and (3–√33)/2.
Problem 2: Find the roots of equation f(x) = 4x3 – 10x2 + 4x = 0.
Solution:
Given expression: f(x) = 4x3 – 10x2 + 4x = 0
⇒ x (4x2 - 10x + 4) = 0
⇒ x (4x2 - 8x - 2x + 4) = 0
⇒ x(4x(x - 2) - 2(x - 2)) = 0
⇒ x (4x - 2) (x - 2) = 0
⇒ x = 0 or 4x - 2 = 0, x - 2 = 0
⇒ x = 0 or x = 1/2 or x = 2
Hence, the roots of the given equation are 0, 1/2 and 2.
Problem 3: Find the roots of equation f(x) = x3 + 3x2 + x + 3 = 0.
Solution:
Given expression: f(x) = x3 + 3x2 + x + 3 = 0.
⇒ x2 (x + 3) + 1(x + 3) = 0
⇒ (x + 3) (x2 + 1) = 0
⇒ x + 3 = 0 or x2 + 1 = 0
⇒ x = -3, ±i
So, the given equation has a real root, i.e., -3, and two imaginary roots, i.e., ±i.
Problem 4: Find the roots of equation f(x) = x3 – 7x2 – x + 7 = 0.
Solution:
Given expressions,
f(x) = x3 – 3x2 – 5x + 7 = 0
First, factorize the equation, f(x): x3 – 3x2 – 5x + 7= 0
It can be factored into (x-7)(x+1)(x-1) = 0
After factoring the polynomial, we can find the roots by equating each factor to zero. For example:
- x - 7 = 0, so x = 7
- x + 1 = 0, so x = -1
- x - 1 = 0, so x = 1
So the roots of the equation f(x): x3 – 3x2 – 5x + 7 = 0 are
Problem 5: Find the roots of equation f(x) = x3 − 6x2 + 11x − 6 = 0, using the graphical method.
Solution:
Given expression: f(x) = x3 − 6x2 + 11x − 6 = 0.
Now, simply substitute random values for x in the graph for the given function:
We can see that the graph has cut the X-axis at 3 points, therefore, there are 3 real solutions.
From the graph, the solutions are: x = 1, x = 2, and x = 3.
Hence, the roots of the given equation are 1, 2, and 3.
Practice Problems on Solving Cubic Equations
Various practice problems related to cubic equations are added below. Solve these problems to fully grasp the concept of How to Solve the Cubic equation.
- Question 1: Solve the cubic equation, 3x3+ 2x2 - 11x + 7 = 0.
- Question 2: Find the roots of the cubic equation, 4x3- 12x2 + 17 = 0.
- Question 3:Solve the cubic equation, x3+ 4x2 - x + 3 = 0 using graphical method.
- Question 4: Find the number which satisfies, -9x3+ 11x2 - 8x + 2 = 0.
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