Factoring third degree polynomials

In the conceptualization of algebraic expressions, we observe that polynomials are the functions that have three or more added terms. These can be represented graphically and algebraically. Third degree polynomials are also known as cubic equation or third degree equations. Its general expression has the form:

ax³ + bx² + cx + d = 0

Where a, b, c y d are integers, x the unknown variable of the equation. Finding its value, we will find the possible solutions to this equation or the roots of the polynomial. Let us remember that factorization consists of transforming an algebraic expression, to an expression of multiplication of terms.

Next we will solve polynomials of third degree, by the factorization method.

Example 1: Find the set of solutions of the following cubic equation  x³ - 8x² + x - 8 = 0.

To solve it we will use the factorization:

x² (x - 8) + (x - 8) = 0

We obtain the common factor (x - 8):

(x - 8) (x² + 1) = 0

Let's look at the expression, so that such equality can be fulfilled, (x - 8 = 0) or (x² + 1 = 0). Let's see each one separately:

x - 8 = 0

x = 8

So one of the solution values to this equation is x = 8. Let's see the following expression:

x² + 1 = 0

x² = - 1

Remember that the root of a negative number is an imaginary number (√-1 = i).

x = ± √-1

We have two solutions:

x1 = i

x2 = -i

So the solution roots to this equation are:

{8, -i, i}

{8} Solution in real numbers, {i, -i} for imaginary numbers.

We will check the solution for the real numbers: We substitute x = 8 in the initial expression of the exercise x³ - 8x² + x - 8 = 0:

(8) ³ - 8 (8) ² + 8 - 8 = 0

Recall the power property na× nb= n(a + b).

(8) ³ - (8) ³ + 0 = 0

0 + 0 = 0

0 = 0

This equality is true, therefore x = 8 is a solution to the third degree equation in real numbers. Graphical representation of the polynomial:

factoring polynomials example

Example 2: Find the solutions of the following cubic equation x³  - 8 = 0.

To find the roots of this polynomial we must remember how to solve the remarkable product of the cube term differences:

a³ - b³ = (a - b) (a² + ab + b²)

We can rewrite the equation:

x³ - 8 = 0

x³ - 2³ = 0

(x - 2) (x² + 2x + 4) = 0

For this equality to be fulfilled, (x - 2 = 0) or (x² + 2x + 4 = 0); Let's look at the equations separately, first:

x - 2 = 0

x = 2

A solution to this cubic equation is x = 2.

Second equation:

x² + 2x + 4 = 0

Since we cannot factor it, we will use the solver x = [- b ± √ (b² - 4ac)] / 2a. Where a = 1, b = 2 and c = 4 we substitute these values in the resolver and we have:

x = [- 2 ± √ (2² - 4 · 1 · 4)] / (2 · 1)

x = [- 2 ± √ (-12)] / 2

x = [- 2 ± i √ (3 · 4)] / 2

x = (- 2 ± i 2√3) / 2

x = - 1 ± i √3

So the solution roots to this equation are:

{two ; -1 + i √3; -1 - i √3}

{2} is the solution in real numbers and {-1 + i √3; -1 - i √3} for imaginary numbers.

We will check the solution for the real numbers:

We substitute x = 2 in the initial expression of exercise x³ - 8 = 0:

(2) ³ - 8 = 0

8 - 8 = 0

0 = 0

This equality is true therefore x = 2 is a solution to the third degree equation in real numbers. The polynomial graphically:

factoring polynomials2