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This article will show you how to factor a 3rd degree polynomial. We will learn how to factorize using the common factorization method and the method of using the free terms.

## Steps

### Factoring using the group method

**Divide the polynomial into two groups.**We need to divide the polynomial into two groups and “treat” each group separately.

- Assume “consider polynomials.” x
^{3}+ 3x^{2}– 6x – 18 = 0. We group the polynomial into two parts (x^{3}+ 3x^{2}) and (- 6x – 18).

**Find the common factor of each group.**

- In the group (x
^{3}+ 3x^{2}), we can easily see that x^{2}is a common factor. - In the group (- 6x – 18), -6 is the common factor.

**Draw the common factor out of each group.**

- Draw x
^{2}as the common factor of the first group, we get x^{2}(x + 3). - Draw -6 out of the second group, we get -6(x+3).

**If a factor occurs in both groups, we can group the two groups together.**

- We have (x + 3)(x
^{2}– 6).

**Find test.**In case we have a factor containing x

^{2}, it should be noted that

*both*positive and negative values obtained from this factor satisfy the equation.

- The roots of the polynomial under consideration are -3, √6 and -√6.

### Factoring using the free term

**Rearrange the polynomial to the form aX**

^{3}+bX^{2}+cX+d.- For example, consider the formula x
^{3}– 4x^{2}– 7x + 10 = 0.

**Find all the factors of “d”.**The constant “d” is the zero associated with any variable, in this case the variable is “x”.

- Multipliers of a number are numbers that we can multiply by another number to get another number. In this case, the factors of 10, or “d,” are: 1, 2, 5, and 10.

**Find a factor that can make the polynomial equal to 0.**We want to determine the factor that when this factor is substituted into the variable “x”, the polynomial will be 0.

- Try with the first factor, 1. Substitute “1” into all the “x” variables in the polynomial:

(1)^{3}– 4(1)^{2}– 7(1) + 10 = 0 - We get: 1 – 4 – 7 + 10 = 0.
- Since 0 = 0, we have x = 1 as a solution of the equality.

**Change position.**If x = 1, we can rearrange the equality slightly without changing its meaning.

- “x = 1” is equivalent to “x – 1 = 0” or “(x – 1)”. That is, we have subtracted 1 on both sides of the equation.

**Separate the solution from the rest of the equation.**“(x – 1)” is the solution. Try separating this from the equation and see if that helps. Proceed with the polynomials one by one.

- Can we split (x – 1) from x
^{3}? The answer is no. However, we can borrow -x^{2}from the second variable and do the factorization as follows: x^{2}(x – 1) = x^{3}– x^{2}. - Can we separate (x – 1) from the rest of the second variable? Again the answer is no. We need to borrow part of the third variable. Taking 3x from -7x and grouping the common factor with the rest of the second variable, we get -3x(x – 1) = -3x
^{2}+ 3x. - Since we borrowed 3x from -7x, so the second variable will become -10x, notice the free term is 10. Can we factor it? The answer is yes: -10(x – 1) = -10x + 10.
- We have separated and rearranged the variables so that we can group (x – 1) as a common term for the whole expression. In general, we have the expression after splitting x
^{3}– x^{2}– 3x^{2}+ 3x – 10x + 10 = 0, this expression is also equivalent to x^{3}– 4x^{2}– 7x + 10 = 0.

**Continue the substitution of the free term.**Consider the numbers that have been split when drawing (x – 1) as a common factor in step 5:

- x
^{2}(x – 1) – 3x(x – 1) – 10(x – 1) = 0. We can rearrange this equality to make it easier to factorize: (x – 1)(x^{2}– 3x – 10) = 0. - At this point, we need to factorize for (x
^{2}– 3x – 10). This expression can be decomposed into (x + 2)(x – 5).

**The answer to the equation is the extracted solution.**We can check if the result obtained is exactly the solution of the equation by substituting the found value into the variables of the original polynomial.

- (x – 1)(x + 2)(x – 5) = 0, that is, 1, -2 and 5 are the roots of the polynomial.
- Substituting -2 into the original equation we get: (-2)
^{3}– 4(-2)^{2}– 7(-2) + 10 = -8 – 16 + 14 + 10 = 0. - Substituting 5 into the original equation, we get (5)
^{3}– 4(5)^{2}– 7(5) + 10 = 125 – 100 – 35 + 10 = 0.

## Advice

- For real numbers, there is no third degree polynomial that cannot be factored because all cubics have at least one real root. For polynomials that do not have a suitable real root, for example x^3 + x + 1, we cannot decompose into polynomials if we use real numbers with rational coefficients. Although it is possible to calculate the solution of this polynomial according to the cubic equation solution formula, it cannot itself be decomposed into
*integer*polynomials. - A third degree polynomial is the product of three first degree polynomials or the product of a first degree polynomial and a second degree polynomial that cannot be factored. In this case, after we have found the first common factor, we can do polynomial division by the polynomial to find the quadratic polynomial.

wikiHow is a “wiki” site, which means that many of the articles here are written by multiple authors. To create this article, 24 people, some of whom are anonymous, have edited and improved the article over time.

This article has been viewed 130,925 times.

This article will show you how to factor a 3rd degree polynomial. We will learn how to factorize using the common factorization method and the method of using the free terms.

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