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Factoring

What is factoring?

Factoring is the process of decomposing or splitting any given polynomial into a product of two or more polynomials. We always do this with numbers. For example, here are some possible ways to factor 24.

24 = 1 x 2424 = 2 x 1224 = 3 x 8
24 = 4 x 624 = $\frac{1}{2}$ x 4824 = -2 x -12
24 = -4 x -624 = -3 x -824 = 2 x 2 x 2 x 2 x 3

The process of factoring or decomposing the factors of polynomials is called factorization. The factors of a polynomial should always be less than the degree or equal to the original polynomial. 

What are the different ways to factor polynomials?

In this section, we will learn the three different methods to factor any polynomials. 

Factoring by greatest common factor (GCF)

The first and simplest method that we are going to talk about is factoring out the greatest common factor (GCF) of the terms in the polynomial. In general, this should be the first thing that you should try because it often simplifies a complex polynomial. 

Here are some of the strategies and techniques you can try in finding the greatest common factor of any polynomials.

  1. Completely factor out each term.
  2. Write a product using factors that is common to all the terms.

Remember that factoring out the GCF is like using the distributive law in reverse. The distributive law states that if 

a(b + c) = ab + ac

However, we simply use this law in reverse when we factor out the greatest common factor. Hence, if

ab + ac = a(b + c)

Notice that each term in the polynomial ab + ac has an “a”, so we can factor it out. 

Example #1

Factor out the greatest common factor (GCF) of the polynomial x4 – x3 – x2 + x.

Solution

Step-by-step ProcessExplanation
x4 – x3 – x2 + xObserve the given polynomial and see what each term have in common. In this case, each term contains an x
x4 = x3 .x
-x3 = -x2 .x
-x2 = -x. x
x = 1 x
Factor out x in each term.
(x)(x3 – x2 – x + 1)Write the result when x is already factored out in each term. 
Therefore, the result of factoring out the greatest common factor of the polynomial x4 – x3 – x2 + x is (x)(x3 – x2 – x + 1) where x is the GCF.

Example #2

Factor out the greatest common factor (GCF) of the polynomial 3x4 + 6x3 – 18x2 + 9x.

Solution

Step-by-step ProcessExplanation
3x4 + 6x3 – 18x2 + 9xObserve the given polynomial and see what each term has in common. In this case, each term contains a product of a number and x.  
3x4 = 3 x4
6x3 = 3 2 x3
-18x2 = 3 -6 x2
9x = 3 . 3 x
Since each term contains a number that is a multiple of 3, we can simply factor out 3 in each term first. 
(3)(x4 + 2x3 – 6x2 + 3x)Write the result when 3 is already factored out. 
x4 = x x3
2x3 = x 2x2
-6x2 = x -6x
3x = x 3
Observe that we can still factor out x in each term in the polynomial x4 + 2x3 – 6x2 + 3x. Hence, factor out x in each term.
3 x = 3xGet the product of the common factors. The common factors of each term are 3 and x
(3x)(x3 + 2x2 – 6x + 3)Write the final result of factoring out 3x in each term.
Therefore, the result of factoring out the greatest common factor of the polynomial 3x4 + 6x3 – 18x2 + 9x is (3x)(x3 + 2x2 – 6x + 3) where 3x is the GCF.

Example #3

Factor out the greatest common factor (GCF) of the polynomial x5y4 – x2y3 + xy2.

Solution

Step-by-step ProcessExplanation
x5y4 – x2y3 + xy2Observe the given polynomial and see what each term has in common. Since there exists x and y in each term, we can factor out the x’s and y’s. 
x5y4 = xy2 x4y2
-x2y3 = xy2 -xy
xy2 = xy2 1
Get the smallest degree of x and y. By looking at each term, we can factor out xy2 and apply the laws of exponent.
(xy2)(x4y2 – xy + 1)Write the final result of factoring out xy2 in each term.
Therefore, the result of factoring out the greatest common factor of the polynomial x5y4 – x2y3 + xy2 is (xy2)(x4y2 – xy + 1) where xy2 is the GCF.

Example #4

Factor out the greatest common factor of the polynomial (5x7)(3x +2) + (25x5)(3x +2).

Solution

Step-by-step ProcessExplanation
5x7(3x +2) + 25x5(3x + 2)Observe the given polynomial and see what each term has in common. We can easily see that 3x +2 is common to each term. 
5x7(3x +2) = (3x + 2) 5x7
25x5 = (3x + 2) 25x5
Factor out 3x + 2 from each term. 
(3x + 2)(5x7 + 25x5)Write the result of factoring out 3x + 2 from the given polynomial. 
5x7 = 5x5 x2
25x5 = 5x5 5
Observe that the terms 5x7 + 25x5 still have common factors.  Hence, we can factor out 5x5 from it. 
(3x + 2)(5x5)(x2 + 5)Write the final result of factoring out 3x + 2 and 5x from the given polynomial.
Therefore, the result of factoring out the greatest common factor of the polynomial (5x7)(3x +2) + (25x5)(3x +2) is (3x + 2)(5x5)(x2 + 5) where (3x + 2)(5x5) is the GCF.

Factoring by groupings

Factoring by groupings is done when no common factor exists to all of the terms of a polynomial, but there are factors common to some of its terms. Hence, our main goal here is to find groups with common factors. 

Example #1

Given the polynomial 5x2 + 9x – 10x – 18, factor out using the method of groupings.

Solution

Step-by-step ProcessExplanation
5x2 + 9x – 10x – 18Observe the given polynomial and see if each term has a common factor. Since not all terms have a common factor, we can use factorization by grouping. 
(5x2 – 10x) + (9x – 18)Group the terms that have common factor.  
5x2 = 5x x
-10x = 5x -2
Factor out 5x from 5x2 and -10x.
5x(x – 2) + (9x – 18)Write the result of factoring out 5x from the two terms. 
9x = 9 x
18 = 9 -2
Factor out 9 from 9x and -18. 
5x(x – 2) + 9(x – 2)Write the result of factoring out 9 from the two terms.
(x – 2)(5x + 9)Factor out x – 2.
Therefore, the result of factoring by groupings of 5x2 + 9x – 10x – 18 is (x – 2)(5x + 9).

Example #2

Using factorization by groupings, factor out the polynomial 6x2 + 14x + 9x + 21. 

Solution

Step-by-step ProcessExplanation
6x2 + 14x + 9x + 21Observe the given polynomial and see if each term has a common factor. Since not all terms have a common factor, we can use factorization by grouping. 
(6x2 + 9x) + (14x + 21)Group the terms that have common factor.  
6x2 = 3x 2x
9x = 3x 3
Factor out 3x from 6x2 and 9x. 
3x(2x + 3) + (14x + 21)Write the result of factoring out 3x from the two terms.
14x = 7 x 2x
21 = 7 x 3
Factor out 7 from 14x and 21.
3x(2x + 3) + 7(2x +3)Write the result of factoring out 7 from the two terms.
(2x + 3)(3x + 7)Factor out 2x +3.
Therefore, the result of factoring by groupings of 6x2 + 14x + 9x + 21 is (2x + 3)(3x + 7).

Example #3

Factor out the polynomial x7 + x5 – 10x4 – 10x2 using factorization by groupings.

Solution

Step-by-step ProcessExplanation
6x2 + 14x + 9x + 21Observe the given polynomial and see if each term has a common factor. Since not all terms have a common factor, we can use factorization by grouping. 
(6x2 + 9x) + (14x + 21)Group the terms that have common factor.  
6x2 = 3x 2x
9x = 3x 3
Factor out 3x from 6x2 and 9x. 
3x(2x + 3) + (14x + 21)Write the result of factoring out 3x from the two terms.
14x = 7 2x
21 = 7 x 3
Factor out 7 from 14x and 21.
3x(2x + 3) + 7(2x +3)Write the result of factoring out 7 from the two terms.
(2x + 3)(3x + 7)Factor out 2x +3.
Therefore, the result of factoring by groupings of 6x2 + 14x + 9x + 21 is (2x + 3)(3x + 7).

Factoring quadratic polynomials

Quadratic polynomial is a polynomial where the highest degree of a term is in degree two. When factoring quadratic polynomials, we are simply doing the FOIL method in reverse. Here are some things that will help you factor quadratic polynomials.

  1. Arrange the given polynomial in a standard form, ax2 + bx + c.
  2. List all the possible factors of c
  3. Make sure that the sum of the factors of c is the same as b
  4. If the last term is negative, it will have the form (x + __)(x – __).
  5. If the second term is negative, it will have the form (x – __)(x – __).
  6. If all terms of the polynomial is positive, it will be in the form (x + __)(x + __). 
  7. If all terms of the polynomial is negative, it will be in the form –(x + __)(x + __). 

Example #1

Factor the quadratic polynomial x2 + 5x + 6. 

Solution

Step-by-step ProcessExplanation
x2 + 5x + 6Observe the given polynomial.
(x + __)(x + __)Since we have the leading term x2, there is only one way on how we can split it. x2 = x xHence, we are certain that the first term of each factor is x.     
6 = 2 3
6 = -2 -3
Now, let’s focus on 6 and list all its possible factors. Since we are looking for factors of 6 that will have the sum of 6, we will use the factors 2 and 3.  
(x + 3)(x + 2)Plug in the numbers 2 and 3. 
Therefore, the result of factoring out the quadratic polynomial x2 + 5x + 6 is (x + 3)(x + 2).

Example #2

What is the factored form of the quadratic formula x2 + x – 20?

Solution

Step-by-step ProcessExplanation
x2 + x – 20Observe the given polynomial.
(x + __)(x – __)The leading term is x2, which means we can easily write its factor as x x. More so, since the last term is negative, we will use the form (x + __)(x – __). 
-20 = 2 -10
-20 = -2 10
-20 = -4 5
-20 = 4 -5
Now, list all the possible factors of -20. Since, we are looking for factors that when added will give us a result of 1, we will use the factors -4 and 5. 
(x + 5)(x – 4)Plug in -4 and 5. 
Therefore, the result of factoring out the quadratic polynomial x2 + x – 20is (x + 5)(x – 4).

Example #3

Factor the quadratic polynomial x2 – 15x + 54. 

Solution

Step-by-step ProcessExplanation
x2 – 15x + 54Observe the given polynomial.
(x – __)(x – __)The leading term is x2, which means we can easily write its factor as x x. Additionally, since the second term is negative, we will use the form (x – __)(x – __).
54 = 2 27
54 = 3 18
54 = 6 9
Now, list all the possible factors of 54. Since, we are looking for factors that will give us the sum of 15, we will use the factors 6 and 9.  
(x – 6)(x – 9)Plug in 6 and 9. 
Therefore, the result of factoring out the quadratic polynomial x2 – 15x + 54is (x – 6)(x – 9).

Factoring using algebraic identities

Using algebraic identities, the process of factorization can be easily done. Below is the table that shows some algebraic identities you can use in factoring.

a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2
a2 – b2 = (a + b)(a – b)
a3 + b3 = (a + b)(a2 – ab + b2)
a3 – b3 = (a – b)(a2 + ab + b2)

Example #1

Factor the polynomial x2 + 14x + 49 using algebraic identities.

Solution

Step-by-step ProcessExplanation
x2 + 14x + 49Observe the given polynomial.
(a + b)2Using algebraic identities, it satisfies the identity a2 + 2ab + b2 = (a + b)2. Hence, we need to find the values of a and b.
a2 = x2
√a2 = √x2
a = x
By getting the square root of a2 and x2, we will get the value of a as x.  
b2 = 49
√b2 = √49
b = 7
Get the square root of b2 and 49. Hence, the value of b is 7. 
(a + 7)2Plug in the values of a and b
Therefore, the result factoring the polynomial x2 + 14x + 49 using algebraic identities is
(a + 7)2 or (a + 7)(a + 7).

Example #2

Factor the polynomial 4x2 – 81.

Solution

Step-by-step ProcessExplanation
4x2 – 81Observe the given polynomial.
(a + b)(a – b)Use the algebraic identitya2 – b2 = (a + b)(a – b)
a2 = 4x2
 √a2 = √4x2
a = 2x
Find the value of a by getting the square root of the first term. 
b2 = 81
√b2 = √81
b = 9
Get the square root of the last term. 
(2x + 9)(2x – 9)Plug in the values of a and b in the form (a + b)(a – b)
Therefore, the result factoring the polynomial 4x2 – 81 using algebraic identities is (2x + 9)(2x – 9).

Example #3

Factor the polynomial 27x3 + 8. 

Solution

Step-by-step ProcessExplanation
27x3 + 8Observe the given polynomial.
(a + b)(a2 – ab + b2)Use the algebraic identitya3 + b3 = (a + b)(a2 – ab + b2)
a3 = 27x3
$\sqrt[3]{a^3} = \sqrt[3]{27x^3}$
a = 3x
Find the value of a by getting the cube root of the first term. 
b3 = 8
$\sqrt[3]{b^3} = \sqrt[3]{8}$
b = 2
Get the cube root of the last term. 
(3x + 2)[(3x)2 – 3x 2 + 22]Plug in the values of a and b in the form (a + b)(a2 – ab + b2)
(3x + 2)(9x2 – 6x + 4)Simplify. 
Therefore, the result factoring the polynomial 27x3 + 8 using algebraic identities is (3x + 2)(9x2 – 6x + 4).

What is the importance of factoring polynomials? 

Factoring is a vital knowledge and fundamental step that helps us easily understand equations. Every time we rewrite complex polynomials into a simpler polynomials, we apply the concept of factoring – hence, giving us more information about the components of the equation or algebraic expressions.

Factoring and Expanding Linear Expressions with Rational Coefficients 7th Grade Math Worksheets
GCF: Greatest Common Factor (Valentine’s Day Themed) Math Worksheets
Factors and Multiples (Ages 8-10) Worksheets (Space themed)

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