# Factoring a 3rd Degree Polynomial – Methods and Examples

Factoring polynomials helps us determine the zeros or solutions of a function. However, factoring a 3rd-degree polynomial can become more tedious. In some cases, we can use grouping to simplify the factoring process. In other cases, we can also identify differences or sums of cubes and use a formula.

We will look at both cases with examples.

##### ALGEBRA

Relevant for

Learning to factor 3rd degree polynomials with examples.

See methods

##### ALGEBRA

Relevant for

Learning to factor 3rd degree polynomials with examples.

See methods

## Factor 3rd degree polynomials by grouping

Grouping methods can simplify the process of factoring complex polynomials. Analyzing the polynomial, we can consider whether factoring by grouping is feasible.

If the polynomial is in a form where we can remove the greatest common factor of the first two terms and the last two terms to reveal another common factor, we can employ the grouping method by following these steps:

Step 1: Group the polynomial into two parts. By grouping the polynomial into two parts, we can manipulate these parts individually.

For example, if we want to factor the polynomial $latex {{x}^3}+2{{x}^2}-4x-8$, we can group it into $latex ({{x}^3}+2{{x}^2})$ and $latex (-4x-8)$.

Step 2: Find the common factor in each part.

In the part $latex ({{x}^3}+2{{x}^2})$, we see that $latex {{x}^2}$ is a common factor.

In the $latex (-4x-8)$ part, we see that -4 is a common factor.

Step 3: Factor the common factors of both terms.

Factor the $latex {{x}^2}$ from the first part to obtain $latex {{x}^2}(x+2)$.

Factor the -4 from the second part to get $latex -4(x+2)$.

Step 4: If each of the two terms contains the same factor, we can combine them.

This results in $latex (x + 2)({{x}^2}-4)$.

Step 5: If possible apply the difference of squares.

In this case, we can apply the difference of squares in the expression $latex ({{x}^2}-4)$ to obtain $latex (x+2)(x+2)(x-2)$.

We have already obtained the factored expression.

### EXAMPLE 1

Factor the polynomial $latex {{x}^3}-{{x}^2}-4x+4$.

Solution: When we remove the greatest common factor from the first two and the last two terms, we get the following:

$latex {{x}^2}(x-1)-4(x-1)$

Now, we can factor el $latex (x-1)$ from each part to obtain:

$latex ({{x}^2}-4)(x-1)$

Applying the difference of squares, we obtain:

$latex (x+2)(x-2)(x-1)$

### EXAMPLE 2

Factor the polynomial $latex 45{{x}^3}+18{{x}^2}-5x-2$.

Solution: We can group and factor as follows:

$latex (45{{x}^3}+18{{x}^2})-(5x+2)$

$latex 9{{x}^2}(5x+2)-1(5x+2)$

$latex (9{{x}^2}-1)(5x+2)$

We can apply the difference of squares to the first factor:

$latex (3x+1)(3x-1)(5x+2)$

## Factor differences of cubes

If the polynomial has only two terms, each with a perfect cube, we can factor it based on known cubic formulas.

An expression of the form $latex {{a}^3} – {{b}^3}$ is called a difference of cubes. The factored form of $latex {{a}^3} – {{b}^3}$ is $latex (a-b)(a^2+ab+{{b}^2})$. We can check this as follows:

$latex (a-b)({{a}^2}+ab+{{b}^2})$

$latex ={{a}^3}-{{a}^2}b+{{a}^2}b-a{{b}^2}+a{{b}^2}-{{b}^3}$

$latex {{a}^3}-{{b}^3}$

### EXAMPLE 1

Facto the polynomial $latex 8{{x}^3}-125$.

Solution: In this case, we have $latex a=2x$ and $latex b=5$, therefore, the factored form is:

$latex (2x-5)(4{{x}^2}+10x+25)$

## Factor sums of cubes

An expression of the form $latex {{a}^3}+{{b}^3}$ is called a sum of cubes. The factored form of $latex {{a}^3}+{{b}^3}$ is $latex (a+b)({{a}^2} -ab + {{b}^2})$. We can check this as follows:

$latex (a+b)({{a}^2}-ab+{{b}^2})$

$latex ={{a}^3}+{{a}^2}b-{{a}^2}b-a{{b}^2}+a{{b}^2}+{{b}^3})$

$latex ={{a}^3}+{{b}^3}$

### EXAMPLE 1

Factor the polynomial $latex 64{{x}^3}+125$.

Solution: In this case, we have $latex a=4x$ y $latex b=5$, therefore, the factored form is:

$latex (4x+5)(16{{x}^2}-20x+25)$

## How to solve a 3rd-degree polynomial?

To solve a 3rd-degree polynomial, we have to start by factoring the polynomial with any of the factoring methods seen above. If we have a sum of perfect cubes, we use the formula $latex {{a}^3}+{{b}^3}=(a+b)({{a}^2}-ab+{{b}^2})$.

If we have a difference of perfect cubes, we use the formula $latex a^3-{{b}^3}=(a-b)({{a}^2}+ab+{{b}^2})$. In other cases, we can use the grouping method.

After obtaining the factors of the polynomials, we can set each factor equal to zero and solve individually. For example, in an example we saw earlier, we found that the factorization of $latex {{x}^3}-{{x}^2}-4x+4$ is $latex (x+2)(x-2)(x-1)$.

Then, equating each of these factors with zero, we have:

$latex (x+2)=0$   ⇒   $latex x=-2$

$latex (x-2)=0$   ⇒   $latex x=2$

$latex (x-1)=0$   ⇒   $latex x=1$

Therefore, the roots of the polynomial are $latex x=-2, x=2, x=1$.