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How To Solve Cubic Equation Ax^3+ Bx^2+ Cx+D = 0

Cubic Equation

Are you a mathematics student stuck at a cubic equation? Well, facing your high school mathematics at the university can be a little embarrassing because mathematics and students are enemies of long. And I can see that you are also one of those contenders who is poking this subject. My Assignment Services is going to help you take it down.

One step at a time here is everything that you should know about solving a cubic equation.

Returning to the roots, literally.

Mathematics terms can be really tricky, especially for the cubic equations. Every equation has a number of solution values for x which are also called roots. Greater the power of the equation, greater the number of real roots.

The general mathematical representation of a cubic equation is ax3+ bx2+ cx+d = 0.

The coefficients a, b, c, and d can either be a real number or a complex number where a is not equal to 0. Why? Because the equation must have an x3 term in order to be cubic. Except ‘a’, any other coefficient can be equal to 0.

Some cubic equations for example.

  1. 4x3+2x2-11x+12 = 0
  2. 13x3-3x2-98 = 0
  3. 34x3+76x2-29x = 0
  4. 2x3+23x2 = 0
  5. 12x3-48x = 0
  6. 45x3-360 = 0

How to solve these cubic equations?

If you remember, a quadratic may or may not have any real root. But the cubic equation is a little better than that. No, it does not mean that cubic equations always have all real solution. Instead, the cubic equations will always have at least one real root.

Pro tip 101 – To solve a cubic equation, always try to reduce it down to a quadratic equation. That means, reducing the equation to the one where the maximum power of the equation is 2. Then, solve the equation by either factorising or using the quadratic formula.

Always try that your cubic equation is arranged in the general form of the cubic equation which is ax3+ bx2+ cx+d = 0.

Let us take an example to understand the process easily.

Example 1. Solving x2+4x-1 = 6/x

Step 1 – General Form

As you can see, the equation is not in the general form of the cubic equation. So, our first step will be to arrange the equation in the general form of the cubic equation.

To do that, multiply the entire equation with x to eliminate the one in the factor. Through this, your equation becomes x3 as well and will look like –

x3+4x2-x = 6

Step 2 – 0 on the Right Hand Side

You need to subtract 6 from both sides of the equation so that you have a 0 on the right-hand side. After doing that, you will have –

x3+4x2-x-6 = 0

To Solve These Questions, Use Factor Theorem

Now you must be wondering what in lord’s name is a factor theorem, aren’t you? Factor theorem states that a polynomial p(x) has a factor (x – a) if and only if f(a)=0 (i.e. a) is a root.

Did not catch that one? Okay, let me break it down.

Imagine there is a polynomial p(x) x3+x2+x+1 = 0. There is a root for the equation, suppose 2. So, the factor theorem says that if 2 is the root of this equation, then (x-2) will be a factor of the polynomial.

Hence, the polynomial can be written as –

p(x) = (x – a)q(x) + r(x)

Since remainder is 0, the polynomial reduces to

p(x) = (x – a)q(x)

Solve x3–5x2–2x+24 = 0

Lets say that x = -2 is a solution of this polynomial. So, according to the factor theorem, (x+2) becomes a factor of this polynomial.

Hence, we can rewrite x3–5x2–2x+24 = 0 as –

(x+2)(x2+ax+b) = 0

a and b are coefficients whose value you need to find using the synthetic division method.

Step 1

Identify the coefficients of the original cubic equation. They are 1, -5, -2 and 24.

1  -5 -2  24 x = -2
1

Step 2

Now multiply number (1) that just brought down by the known root -2. As a result is -2, you mention the result in the other line.

1  -5 -2  24

-2

x = -2
1

Step 3

Add the numbers in the second column which gives the result –

1  -5 -2  24

-2

x = -2
1  -7

Step 4

Multiply the sum with the known root, i.e. -7 multiplied by -2.

The result thus obtained will be written in the second row.

1  -5 -2  24

-2  14

x = -2
1  -7

Step 5

Add the numbers in the third column.

1  -5 -2  24

-2  14

x = -2
1  -7 12

Step 6

Repeating the multiplication process again, we get 12 multiplied by -2.

1  -5 -2  24

-2  14 -24

x = -2
1  -7 12

Adding the digits in the final column, the result will be –

1  -5 -2   24

-2  14 -24

x = -2
1  -7 12   0

Step 7

Now that we have received 0 with the final coefficient, it is sure that -2 is a root of the polynomial in the question.

By seeing the bottom row, you can form a quadratic equation as –

x2–7x+12 = 0

Hence, the reduced polynomial as per the factor theorem will be –

(x+2)(x2–7x+12) = 0

Step 8

Apply the quadratic factorisation technique here to factorise x2–7x+12. The equation becomes –

(x+2)(x–3)(x–4) = 0

Hence, the roots for the polynomial x3–5x2–2x+24 = 0 are

  1. x = -2
  2. x = 3
  3. x = 4

Another example? Well, then solve x3– 7x-6 = 0

Step 1

Assuming that x = -1 is a solution for this polynomial. Hence, the factor becomes (x+1).                                                                      

Therefore, p(x) = (x+1)(x2+ax+b) = 0

Apply synthetic division method with the coefficients of the original cubic equation. They are 1, 0, -7 and -6.

1  0 -7  -6 x = -1
1

Step 2

Now multiply number (1) that just brought down by the known root -1. As the result is -1, you mention the result in the other line.

1  0 -7  -6

  -1

x = -1
1

Step 3

Add the second column.

1  0 -7  -6

  -1

x = -1
1  -1

Step 4

Multiply -1 with the known root -1 and then write the result in the second row.

1  0 -7  -6

  -1 1

x = -1
1  -1

Step 5

Adding the third column now.

1  0 -7  -6

  -1 1

x = -1
1  -1 -6

Step 6

Multiply -6 with the known root -1.

1  0 -7  -6

  -1 1   6

x = -1
1  -1 -6

Adding the final column –

1  0 -7  -6

  -1 1  6

x = -1
1  -1 -6  0

Step 7

The quadratic equation will be –

x2-x– 6 = 0

The reduced polynomial will be –

(x+1)(x2-x– 6) = 0

Step 8

The factors of the polynomial are –

(x+1)(x–3)(x+2) = 0

Therefore, the roots of the equation are –

  1. x = -1
  2. x = 3
  3. x = -2

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Author: Chloe Kirner

Chloe Kirner is a mechanical engineer by profession but has detailed knowledge about other branches as well which encompasses of chemical, civil, electrical, etc. After completing her PhD from Monash University, she worked as a freelance expert, guiding students through their tasks and assignments. She has contributed papers in international conferences and is an active researcher. She especially enjoys guiding students in their research projects by helping them prepare a proposal and then a report. She has written several articles and blogs for the help of students in answering their engineering assignment queries and is currently an engineering assignment help provider at My Assignment Services.

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