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# 3 Ways to Solve Your Cubic Equations in One Variable with Examples

30 Nov 2023

9 minutes

7842 Views

How many of you remember what you were taught in school? Especially the tricks and formulas in math that almost haunted you in nights. Well, now that during your course you are encountering math again, there could be a possibility that a lot of questions confuse you and you are stuck with some of them. Many of them would be seeming almost unsolvable. Especially the ones that are structured with variables along with integral values. And you start taking the help of Free Mathematics Software.

There is no difficulty when we talk about linear equations (the ones with one variable), and quadratic equations (the ones with two variables) too can be solved, but when it comes to solving cubic equations (the ones with three variables), your brain needs to exercise a lot. A lot of students often ask us how to solve cubic equations easily. They ask us about different methods to solve the questions in their assignments. Our experts have listed a few methods that can help you solve a cubic equation in your assignments without panicking.

Before discussing how to find roots of cubic equation easily, let us discuss the difference between the three different equations that have been discussed above.

• Linear Equation: One degree equations: Highest power of the variable is “1” : General Form ax+b=0
• Quadratic Equation: Two degree equations: Highest power of the variable is “2” : General Form ax2+bx+c=0
• Cubic Equation: Three degree equations: Highest power of the variable is “3” : General Form ax3+bx2+cx+d=0

Students find it much easier to work on linear and quadratic equations, but working on cubic equations is an arduous task. They often rush to our math assignment help experts and ask how to solve 3 degree equations in easy steps.

### Here Are the Different Ways to Solve a Cubic Equation

You can easily solve the quadratic equation. When you start solving a cubic equation, all you need to do is, just change your cubic equation to a quadratic equation, put the formula and move.
Sounds Easy!!
To use quadratic formula, it is important that your equation is of the form : ax3+bx2+cx=0

Let the equation be x3+6x2+5x=0

Our equation can be simplified to,
x3+6x2+5x=0
0r x(x2+6x+5)=0
Thus, “0” becomes a root of the equation.
For the other two roots, let’s put the values to the formula and get the results.

For our equation a=1, b=6, c=5
x = [6 ± √(62ï¼(4*1*5))]/2*1
x= [6± √(16)]/2
x= [6±4]/2
Thus, x can take two values, x=5,1.

The roots of our equation are x=0,1,5.

### 2. The Factor Listing Methods

Now let us see a method that can be implemented if we have a value for d as well. In the method, we would see how to find solution of cubic equation without
Let the equation be,  x3+6x2+7x+8=6
This could further be simplified to x3+6x2+7x+2=0
Now, compare the equation to the standard form, ax3+by2+cx+d=0

Here, we would first find the factors of a and d in the equation.
Now, a=1 & d=2

Factors of a= Factors of 1= 1
Factors of d= 1,2

Factors of a / Factors of d

= 1,1/2, and include the negatives too.
= 1,1/2ï¼1,ï¼1/2
The roots to our equation is in the list here.

Now, use the synthetic division method to find the solution of the equation as the process below reflects.

1.(ï¼1)*1 = ï¼1
2.6+(ï¼1) = 5
3.(ï¼1)*5 =ï¼5
4.7+(ï¼5) = 2
5.(ï¼1)*2= ï¼2
6.2+(ï¼2) = 0

Since the remainder is “0,”ï¼1 is a root of our equation.

Now that we have a root of the equation, the other two roots can easily be found by solving the quadratic equation that we get after substituting the value of x as (ï¼1) once.
Or we can use the same method to work on the equation with other possible roots.

#### 3. The Discriminant Method

Students often ask our math assignment solver to deliver the documents with steps on how to solve cubic equations. Our experts ensure all the steps are clearly explained and do not confuse the students, especially when it has loads of calculations that are to be done.

As here,
Let us consider the cubic equation x3-3x2+3x=1.

First, we’ll change it to the form ax3+bx2+cx+d= 0
The equation becomes x3-3x2+3x-1=0

For solving your equation with the method, you would be basically working with the coefficients. So, it is advised to list them as, a=1;b= -3;c=3;d=-1
Now lets get to work.
For this method, first of all we need to find the value of “ 0”

Substituting values in the equation above,
0 = (-3)2-3*1*3
= 9-9
= 0
Now we calculate, “ 1”

Substituting values in the equation above,
1 = 2*(-3)3-9*1*(-3)*1+27*12*(-1)
=-54+81-27
= 0
Done.
Next we calculate,

Lets put the values we have calculated, in the above formula.

= (0)2 - 4(0)3) ÷ -27(1)2
= 0
Since we have,  = 0, our equation has one or two roots.
Now we need to calculate one last value that would help us find the roots of our cubic equation,

Substituting the values from the above calculations in our formula,

C =  3√((√((02 - 4*03) + 0/ 2)
=  0

Now to calculate the roots of the equation. we would use the final formula.

Substituting b=(-3), u = (-1 + √(-3))/2, n=1,2 or 3, C=0, and Δ0=0 in the above formula.
Evaluate the value for this and substitute the result in your equation. If the answer is “0,” then you got the root.

Phew...
That’s a lot of calculations!!

### On the Whole

These are some easy approaches that you can adopt to solve cubic equations that have been assigned to you by your math professor. Our experts suggest to be very quick and smart when you adopt any of these methods. Each of the technique has some conditions that are to be satisfied before moving to the solutions. It is important that you decide to work on these assignments with full precision and do not confuse yourself while working on the questions that you are given.

Feeling Confused?

If you are confused about how to solve your cubic equation, then just reach to the assignment experts at Instant Assignment Help and get it done easily.