Introduction to Factoring Expressions
Everything should be made as simple as possible, but not simpler.
—Albert Einstein (1879–1955) GermanAmerican Theoretical Physicist
In this lesson, you'll learn how to factor single–variable and multivariable algebraic expressions.
As we learned earlier, two numbers or variables that are multiplied are called factors. We multiply factors to find a product. When we break down a product into its factors, that's called factoring.
To factor a whole number, such as 6, we list every other whole number that divides evenly into 6. The numbers 1, 2, 3, and 6 all divide evenly into 6. Every number has itself and 1 as factors, because 1 multiplied by any number is that number. Numbers whose only whole factors are 1 and itself are called prime numbers. Numbers that have at least one other factor are called composite numbers.
Tip:
Every whole number is either prime or composite except the number 1, which is considered neither prime nor composite.

Prime Factorization
The prime factorization of a number is a multiplication sentence made up of only prime numbers, the product of which is the number. For example, the prime factorization of 6 is (2)(3), because 2 and 3 are both prime numbers, and they multiply to 6.
The number 7 is a prime number; its only factors are 1 and 7. Every variable has exactly two factors: 1 and itself. The factors of x are 1 and x. If a term has a coefficient, some of that term are its coefficient and its variables (with their exponents). That coefficient can be broken down into its factors, and the variables with their exponents can be broken down into their factors. The term 8x can be factored into (8)(x), because 8x means "8 multiplied by x." 8 can be factored into (2)(4), and 4 can be factored into (2)(2). The factorization f 8x is (2)(2)(2)(x). None of the values in parentheses can be factored any further.
The term 3x^{2} can be factored into (3)(x^{2}). 3 is a prime number, so it cannot be factored any further, butx2 is equal to (x)(x). 3x^{2} = (3)(x)(x).
Factoring an Expression
Now that we know how to factor a single term, let's look at how to factor an expression. (5x + 10) is made up of two terms, 5x and 10. When we are breaking down an expression of two or more terms, we say that we are looking to "factor out" any numbers or variables that divide both terms evenly. 10 is divisible by 2, but 5x is not, so we cannot factor out 2. 5x is divisible by x, because 5x = (5)(x), but 10 is not divisible by x. 10 and 5x are both divisible by 5, however. We can divide each term by 5: 5x ÷ 5 = x and 10 ÷ 5 = 2. We write 5 outside the parentheses to show that both terms inside the parentheses must be multiplied by 5:5(x + 2.)
Tip:
To check that you've factored correctly, multiply your factors. You should get back your original expression.

Only one of the two terms in 5x + 10 contained the variable x, so we were not able to factor out x If every term in an expression contains the same variable, you can factor out that variable. Sometimes, you can factor out more than one of the same variable.
Every term in the expression (x^{5} + x^{3}) has x in it. The first term, x^{5}, can be factored into (x)(x)(x)(x)(x). The second term, x^{3}, can be factored into (x)(x)(x). Both terms have x as a common factor. In fact, both terms have three x's as common factors. We can factor out three x's, or x3, from each term: (x^{5} + x^{3}) = x^{3}(x^{2} + 1).
Let's look closely at how each term was factored. When we factored x^{3} out of x^{5}, we were left with x^{2}. Remember: the exponent of a base is the number of times that base is multiplied by itself. Every time we factor an x out of x^{5}, its exponent decreases by 1. Because we factored three x's out of x^{5}, we are left with two x's, or x^{2}.
The second term in (x^{5} + x^{3}) is x^{3}. Whenever we factor out an entire term, the number 1 is left behind in its place, because 1 times any quantity is that quantity. When we factor a variable out of an expression, the exponent of the factor is equal to the lowest exponent that variable has in the expression. That might sound confusing, so let's look at a few examples:
 We can factor x^{5} out of (x^{10} + 3x^{6} + 2x^{5}), because every term has an x in it, and the lowest value of an exponent of x is 5, in the last term. (x^{10} + 3x^{6} + 2x^{5}) factors into x^{5}(x^{5} + 3x + 2).
 We can factor a out of (10a^{8}– 3a + 7a^{3}), because every term has an a in it, and the lowest value of an exponent of ais 1, in the middle term. (10a^{8}– 3a+ 7a^{3}) factors into a(10a^{7}– 3 + 7a^{2}).
 We can factor g^{9} out of (g^{9}– 4g^{10} + 11g^{13}), because every term has ag in it, and the lowest value of an exponent of g is 9, in the first term. (g^{9}– 4g^{10} + 11g^{13}) factors into g^{9}(1– 4g + 11g^{4}).
If you factor a variable out of every term, the term with the lowest exponent for that variable will no longer contain that variable. In (x^{5} + x^{3}), the second term contained the lowest exponent of x, and when we had finished factoring, the second term no longer contained the variable x: x^{2}(x^{3} + 1). In (g^{9}– 4g^{10} + 11g^{13}), the first term contained the lowest exponent of g, and when we had finished factoring, the first term no longer contained the variable g: g^{9}(1– 4g + 11g^{4}).
Example
Factor 18x^{3} + 12x.
First, find the largest whole number that divides evenly into both coefficients. The factors of 18 are 1, 2, 3, 6, 9, and 18. The factors of 12 are 1, 2, 3, 4, 6, and 12. The greatest common factor is 6. We can factor out a 6 from both terms. Next, look at the variables. Both terms have x in common. The lowest exponent of x is 1, in the second term. We can factor out x from both terms.
To factor 6x out of both terms, divide both terms by 6x: 18x^{3} ÷ 6x = 3x^{2} and 12x ÷ 6x = 2. 8x^{3} + 12x = 6x(3x^{2} + 2). We can check our work by multiplying: 6x(3x^{2} + 2) = (6x)(3x^{2}) + 6x)(2) = 18x^{3} + 12x, so we have factored correctly
Some expressions can only have a whole number factored out, and other expressions can only have a variable factored out. Some expressions, such as 10x + 7, cannot be factored at all. There is no whole number (other than 1) that divides evenly into 10 and 7, and there is no variable common to both terms.
Factoring Multivariable Expressions
Just as with single–variable expressions, a coefficient or variable can only be factored out of a multivariable expression if the coefficient or variable appears in every term. If two variables are common to every term, then two variables can be factored out of the expression.
Example
Factor 9x^{5}y^{2} + 12xy^{3}.
Start with the coefficients. The greatest common factor of 9 and 12 is 3, so 3 can be factored out of the expression.
Next, check to see if each variable exists in each term: x is in both terms and so is y. Both can be factored out of the expression. The smallest exponent of x is 1 (in 12xy^{3}) and the smallest exponent of y is 2 (in 9x^{5}y^{2}).
Divide each term by 3xy^{2}: 9x^{5}y^{2} ÷3xy^{2} = 3x^{4}, and 12xy^{3} ÷3xy^{2} = 4y.
9x5y2 + 12xy3 = 3xy2(3x4 + 4y)
Example
Factor 7x^{6}y^{6}–2x^{5} + 8x^{5}y^{4}.
The greatest common factor of 7, 2, and 8 is 1, so no constant can be factored out of the expression. The variable x is common to every term, so it can be factored out, but the variable y is not in the middle term, –2x^{5}, so it cannot be factored out. The smallest exponent of x is 5, in both the second and third terms, so we can factor x^{5} out of every term Divide each term by x^{5}: 7x^{6}y^{6} ÷ x^{5} = 7xy^{6}, 2x^{5} ÷ x^{5} = 2, and 8x^{5}y^{4} ÷x^{5} = 8y^{4}. 7x^{6}y^{6}– 2x^{5} + 8x^{5}y^{4} =x^{5}(7xy^{6}– 2 + 8y^{4})
Example
Factor 16a^{4}b^{9}c^{4}– 24a^{3}b^{12}c^{6} + 40a^{5}b^{7}c^{5}.
The greatest common factor of the coefficients, 16, 24, and 40, is 8. The variables a, b, and c are common to every term. The smallest exponent of a is 3, the smallest exponent of b is 7, and the smallest exponent of c is 4, so 8a^{3}b^{7}c^{4} can be factored out of every term: 16a^{4}b^{9}c^{4} ÷ 8a^{3}b^{7}c^{4} = 2ab^{2}, 24a^{3}b^{12}c^{6}÷ 8a^{3}b^{7}c^{4} = 3b^{5}c^{2}, and 40a^{5}b^{7}c^{5} ÷ 8a^{3}b^{7}c^{4} = 5a^{2}c.
16a^{4}b^{9}c^{4} – 24a^{3}b^{12}c^{6} + 40a^{5}b^{7}c^{5} = 8a^{3}b^{7}c^{4}(2ab^{2} – 3b^{5}c^{2} + 5a^{2}c)
Find practice problems and solutions at Factoring Expressions Practice Questions.
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