Greatest Common Factor
Factors A number or mathematical symbol that is multiplied by another number or mathematical symbol to form a product. For example, in the equation `4 * 5 = 20`, `4` and `5` are factors. are the building blocks of multiplication. They are the numbers that you can multiply together to produce another number: `2` and `10` are factors of `20`, as are `4` and `5` and `1` and `20`. To factor a number is to rewrite it as a product. `20 = 4 * 5`.
Likewise to factor a polynomialA monomial or the sum or difference of two or more monomials., you rewrite it as a product. Just as any integer can be written as the product of factors, so too can any monomialA polynomial with exactly one term. `4x`, `-5y^2`, and `6` are all examples of monomials. or polynomial be expressed as a product of factors. FactoringThe process of breaking a number down into its multiplicative factors. is very helpful in simplifying and solving equations using polynomials.
A prime factorA factor that only has itself and `1` as factors. is similar to a prime numberA prime number is a natural number with exactly two distinct factors, `1` and itself. The number `1` is not a prime number because it does not have two distinct factors.: it has only itself and `1` as factors. The process of breaking a number down into its prime factors is called prime factorization The process of breaking down a number (or expression) into its prime multiplicative factors. For example, the prime factorization of `12xy` is `2 * 2 * 3 * x * y`. .
Let’s first find the greatest common factor (GCF) The product of the prime factors that two or more terms have in common. The greatest common factor of `xyz` and `3xy` is `xy`. of two whole numbers. The GCF of two numbers is the greatest number that is a factor of both of the numbers. Take the numbers `50` and `30`.
`50 = 10 * 5`
`30 = 10 * 3`
Their greatest common factor is `10`, since `10` is the greatest factor that both numbers have in common.
To find the GCF of greater numbers, you can factor each number to find their prime factors, identify the prime factors they have in common, and then multiply those together.
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Example |
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Problem |
Find the greatest common factor of `210` and `168`. |
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`210=bb2*bb3*5*bb7`
`168=bb2*2*2*bb3*bb7`
GCF `=bb2*bb3*bb7`
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Answer |
`text(GCF) = 42` |
Because the GCF is the product of the prime factors that these numbers have in common, you know that it is a factor of both numbers. (If you want to test this, go ahead and divide both `210` and `168` by `42`: they are both evenly divisible by this number!)
Finding the greatest common factor in a set of monomials is not very different from finding the GCF of two whole numbers. The method remains the same: factor each monomial independently, look for common factors, and then multiply them to get the GCF.
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Example |
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Problem |
Find the greatest common factor of `25b^3` and `10b^2`. |
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`25b^3=bb5*5*bb(b)*bb(b)*b`
`10b^2=bb5*2*bb(b)*bb(b)` |
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GCF `= bb5*bb(b)*bb(b)` |
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Answer |
`text(GCF) = 5b^2` |
The monomials have the factors `5`, `b`, and `b` in common, which means their greatest common factor is `5 * b * b`, or simply `5b^2`.
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Example |
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Problem |
Find the greatest common factor of `81c^3d` and `45c^2d^2`. |
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`81c^3d=bb3*bb3*3*3*bb(c)*bb(c)*c*d`
`45c^2d^2=bb3*bb3*5*bb(c)*bbc*bbd*d` |
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GCF `=bb3*bb3*bbc*bbc*bbd` |
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Answer |
`text(GCF)=9c^2d` |
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Find the greatest common factor of `56xy` and `16y^3`.
A) `8`
B) `8y`
C) `16y`
D) `8xy^3`
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When two or more monomials are combined (either added or subtracted), the resulting expression is called a polynomial. If you can find common factors for each term of the polynomial, then you can factor the polynomial.
As you look at the examples of simple polynomials below, try to identify factors that the terms of the polynomial have in common.
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Polynomial |
Terms |
Common Factors |
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`6x + 9` |
`6x` and `9` |
`3` is a factor of `6x` and `9` |
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`a^2 - 2a` |
`a^2` and `2a` |
`a` is a factor of `a^2` and `2a` |
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`4c^3 + 4c` |
`4c^3` and `4c` |
`4` and `c` are factors of `4c^3` and `4c` |
To factor a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property to rewrite the polynomial in a factored form. Recall that the distributive property of multiplication over additionThe product of a sum and a number is the same as the sum of the product of each addend and the number. For example, `3(4 + 2) = 3(4) + 3(2)`. states that a product of a number and a sum is the same as the sum of the products.
Product of a number and a sum: `a(b + c) = a*b + a * c`. You can say that “`a` is being distributed over `b + c`.”
Sum of the products: `a * b + a * c = a(b + c)`. Here you can say that “`a` is being factored out.”
In both cases, it is the distributive property that is being used.
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Example |
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Problem |
Factor `25b^3 + 10b^2`. |
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`25b^3=bb5*5*bb(b)*bb(b)*b`
`10b^2=bb5*2*bb(b)*bb(b)`
GCF `=bb5*bb(b)*bb(b)=5b^2` |
Find the GCF. From a previous example, you found the GCF of `25b^3` and `10b^2` to be `5b^2`. |
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`25b^3 = 5b^2 * 5b`
`10b^2 = 5b^2 * 2` |
Rewrite each term with the GCF as one factor.
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`5b^2*5b + 5b^2*2` |
Rewrite the polynomial using the factored terms in place of the original terms. |
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`5b^2(5b + 2)` |
Factor out the `5b^2`. |
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Answer |
`5b^2(5b + 2)` |
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The factored form of the polynomial `25b^3 + 10b^2` is `5b^2(5b + 2)`. You can check this by doing the multiplication. `5b^2(5b + 2) = 25b^3+10b^2`.
Note that if you do not factor the greatest common factor at first, you can continue factoring, rather than start all over.
For example:
Factor out `5`: `25b^3 + 10b^2 = 5(5b^3 + 2b^2)`
Then factor out `b^2`: `5b^2(5b+2)`
Notice that you arrive at the same simplified form whether you factor out the GCF immediately or if you pull out factors individually.
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Example |
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Problem |
Factor `81c^3d + 45c^2d^2`. |
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`bb3*bb3*9*bbc*bbc*c*bbd`
`bb3*bb3*5*bbc*bbc*bbd*d`
`bb3*bb3*bbc*bbc*bbd=9c^2d`
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Factor `81c^3d`.
Factor `45c^2d^2`.
Find the GCF.
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`81c^3d = 9c^2d*9c`
`45c^2d^2=9c^2d*5d` |
Rewrite each term as the product of the GCF and the remaining terms. |
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`9c^2d*9c + 9c^2d*5d` |
Rewrite the polynomial expression using the factored terms in place of the original terms. |
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`9c^2d(9c + 5d)` |
Factor out `9c^2d`.
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Answer |
`9c^2d(9c+5d)` |
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Factor `8a^6 - 11a^5`.
A) `88(a^6 - a^5)`
B) `8a(a^5 - 3)`
C) `a^5(a - 1)`
D) `a^5(8a - 11)`
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The distributive property allows you to factor out common factors. However, what do you do if the terms within the polynomial do not share any common factors?
If there is no common factor for all of the terms in the polynomial, another technique needs to be used to see if the polynomial can be factored. It involves organizing the polynomial in groups.
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Example |
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Problem |
Factor `4ab + 12a + 3b + 9` |
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`(4ab + 12a) + (3b + 9)` |
Group terms into pairs. |
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`4ab=bb2*bb2*bba*b` `12a=3*bb2*bb2*bba` GCF `=4a`
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Find the GCF of the first pair of terms.
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`(4a * b + 4a * 3) + (3b + 9)` `4a(b + 3) + (3b+9)` |
Factor the GCF, `4a`, out of the first group. |
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`3b=bb3*b` `9=bb3*3` GCF `=3` |
Find the GCF of the second pair of terms.
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`4a(b + 3)+(3 * b + 3 * 3)` `4a(b + 3) + 3(b+3)` |
Factor `3` out of the second group.
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`4a(b+3)+3(b+3)`
`(b+3)(4a + 3)` |
Notice that the two terms have a common factor, `(b + 3)`.
Factor out the common factor `(b + 3)` from the two terms. |
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Answer |
`(b+3)(4a+3)` |
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Notice that when you factor two terms, the result is a monomial times a polynomial. But the factored form of a four-term polynomial is the product of two binomials.
This process is called the grouping technique. Broken down into individual steps, here's how to do it (you can also follow this process in the example below).
Let’s try factoring a few more four-term polynomials. Notice that in the example below, the first term is `x^2`, and `x` is the only variable present.
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Example |
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Problem |
Factor `x^2 + 2x + 5x + 10`. |
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`(x^2 + 2x) + (5x + 10)` |
Group terms of the polynomial into pairs. |
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`x(x + 2) + (5x+10)` |
Factor out the like factor, `x`, from the first group. |
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`x(x+2)+5(x+2)` |
Factor out the like factor, `5`, from the second group. |
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`(x+2)(x+5)`
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Look for common factors between the factored forms of the paired terms. Here, the common factor is `(x + 2)`. Factor out the common factor, `(x + 2)`, from both terms. The polynomial is now factored. |
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Answer |
`(x+2)(x+5)` |
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This method of factoring only works in some cases. Notice that both factors here contain the term.
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Example |
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Problem |
Factor `2x^2-3x + 8x - 12`. |
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`(2x^2 - 3x) + (8x - 12)` |
Group terms into pairs. |
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`x(2x - 3) + 4(2x-3)` |
Factor the common factor, `x`, out of the first group and the common factor, `4`, out of the second group. |
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`(x + 4)(2x - 3)` |
Factor out the common factor, `(2x - 3)`, from both terms. |
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Answer |
`(x+4)(2x-3)` |
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Example |
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Problem |
Factor `3x^2 + 3x-2x-2`. |
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`(3x^2 + 3x)+(-2x-2)` |
Group terms into pairs. Since subtraction is the same as addition of the opposite, you can write `-2x - 2` as `+(-2x - 2)`. |
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`3x(x + 1)+(-2x-2)` |
Factor the common factor `3x` out of first group. |
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`3x(x + 1)-2(x+1)` |
Factor the common factor of `-2` out of the second group. Notice what happens to the signs within the parentheses once `-2` is factored out. |
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`(x + 1)(3x - 2)` |
Factor out the common factor, `(x + 1)`, from both terms. |
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Answer |
`(x+1)(3x-2)` |
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Factor `10ab + 5b + 8a + 4`.
A) `(2a + 1)(5b + 4)`
B) `(5b + 2a)(4 + 1)`
C) `5(2ab + b + 8a + 4)`
D) `(4 + 2a)(5b + 1)`
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Sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.
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Example |
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Problem |
Factor `7x^2 - 21x + 5x - 5`. |
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`(7x^2 - 21x) + (5x - 5)` |
Group terms into pairs. |
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`7x(x - 3) + (5x-5)` |
Factor the common factor `7x` out of the first group. |
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`7x(x - 3) + 5(x-1)` |
Factor the common factor `5` out of the second group. |
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`7x(x - 3) + 5(x-1)` |
The two groups `7x(x - 3)` and `5(x - 1)` do not have any common factors, so this polynomial cannot be factored any further. |
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Answer |
Cannot be factored |
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In the example above, each pair can be factored, but then there is no common factor between the pairs!
A whole number, monomial, or polynomial can be expressed as a product of factors. You can use some of the same logic that you apply to factoring integers to factoring polynomials. To factor a polynomial, first identify the greatest common factor of the terms, and then apply the distributive property to rewrite the expression. Once a polynomial in `a * b + a * c` form has been rewritten as `a(b + c)`, where `a` is the GCF, the polynomial is in factored form.
When factoring a four-term polynomial using grouping, find the common factor of pairs of terms rather than the whole polynomial. Use the distributive property to rewrite the grouped terms as the common factor times a binomial. Finally, pull any common binomials out of the factored groups. The fully factored polynomial will be the product of two binomials.