2.7: The Greatest Common Factor and Factor by Grouping
We begin this section with definition of a factor of a term. Because \(24 = 2\cdot 12\), both \(2\) and \(12\) are factors of \(24\). Note that \(2\) is also a divisor of \(24\), because when you divide \(24\) by \(2\) you get \(12\), with a remainder of zero. Similarly, \(12\) is also a divisor of \(24\), because when you divide \(24\) by \(12\) you get \(2\), with a remainder of zero.
Suppose \(m\) and \(n\) are integers. Then \(m\) is a factor of a term, \(n\), if and only if there exists another integer \(k\) so that \(n=m \cdot k\).
A factor is also known as a divisor of a term since the term can be divided evenly by the factor, \(\dfrac{n}{m}=k\).
List the factors of \(24\).
Solution
First, list all possible ways that we can express \(24\) as a product of two positive integers:
\[24=1 \cdot 24 \quad \text { or } \quad 24=2 \cdot 12 \quad \text { or } \quad 24=3 \cdot 8 \quad \text { or } \quad 24=4 \cdot 6 \nonumber \]
Therefore, the factors of \(24\) are \(1,2,3,4,6,8,\) and \(24\).
List the factors of \(18\).
- Answer
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\(1,2,3,6,9,\) and \(18\)
List the factors that \(36\) and \(48\) have in common.
Solution
First, list all factors of \(36\) and \(48\) separately, then determine the factors that are in common.
Factors of \(36\) are: \(1, 2, 3, 4, 6, 9, 12, 18, 36\)
Factorss of \(48\) are: \(1, 2, 3, 4, 6, 8, 12, 16, 24, 48\)
Therefore, the common factors of \(36\) and \(48\) are \(1, 2, 3, 4, 6,\) and \(12\).
List the factors that \(40\) and \(60\) have in common.
- Answer
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\(1,2,4,5,10,\) and \(20\)
The greatest common factor of \(a\) and \(b\) is the largest positive number that divides evenly (with no remainder) both \(a\) and \(b\). The greatest common factor of \(a\) and \(b\) is denoted by the symbolism \(\operatorname{GCF}(a, b)\).
Since factors are also divisors, the abbreviation \(\operatorname{GCD}(a, b)\) also represents the greatest common factor of \(a\) and \(b\).
In Example \(\PageIndex{2}\), we listed the common factors of \(36\) and \(48\). The largest of these common factor was \(12\). Hence, the greatest common factor of \(36\) and \(48\) is \(12\), written \(\operatorname{GCF}(36, 48)=12\).
With larger numbers, it can be more difficult to identify the greatest common factor. Prime factorization can be helpful in this situation!
Find the greatest common factor of \(360\) and \(756\).
Solution
Prime factor \(360\) and \(756\), writing your answer in exponential form.
Thus:
\[\begin{array}{l}{360=2^{3} \cdot 3^{2} \cdot 5} \\ {756=2^{2} \cdot 3^{3} \cdot 7}\end{array} \nonumber \]
To find the greatest common factor, list each factor that appears in common and the highest power that appears in common amongst the terms.
In this case, the factors \(2\) and \(3\) appear in common, with \(2\) being the highest power of \(2\) and \(2\) being the highest power of \(3\) that appear in common. Therefore, the greatest common factor of \(360\) and \(756\) is:
\[\begin{aligned} \mathrm{GCF}(360,756) &=2^{2} \cdot 3^{2} \\ &=4 \cdot 9 \\ &=36 \end{aligned} \nonumber \]
Therefore, the greatest common factor is \(\mathrm{GCF}(360,756)=36\).
Note what happens when we write each of the given numbers in the last example as a product of the greatest common factor and a second factor:
\[\begin{array}{l}{360={\color {Red} 36} \cdot 10} \\ {756={\color {Red} 36} \cdot 21}\end{array} \nonumber \]
In each case, note how the second second factors (\(10\) and \(21\)) contain no additional common factors.
Find the greatest common factor of \(120\) and \(450\).
- Answer
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\(30\)
Finding the Greatest Common Factor of Monomials
Example \(\PageIndex{3}\) reveals the technique used to find the greatest common factor of two or more monomials.
To find the greatest common factor of two or more monomials:
- Find the greatest common factor of the coefficients of the given monomials. Use prime factorization if necessary.
- List each variable that appears in common in the given monomials.
- Raise each variable factor that appears in common to the highest power that appears in common among the given monomials.
Find the greatest common factor of \(6 x^{3} y^{3}\) and \(9 x^{2} y^{5}\).
Solution
To find the \(\mathrm{GCF}\) of \(6 x^{3} y^{3}\) and \(9 x^{2} y^{5}\), we note that:
- The greatest common factor of \(6\) and \(9\) is \(3\).
- The monomials \(6x^3y^3\) and \(9x^2y^5\) have the variables \(x\) and \(y\) in common.
- The highest power of \(x\) in common is \(x^2\). The highest power of \(y\) in common is \(y^3\).
Thus, the greatest common factor is \(\mathrm{GCF}\left(6 x^{3} y^{3}, 9 x^{2} y^{5}\right)=3 x^{2} y^{3}\). Note what happens when we write each of the given monomials as a product of the greatest common factor and a second monomial:
\[\begin{array}{l}{6 x^{3} y^{3}={\color {Red} 3 x^{2} y^{3}} \cdot 2 x} \\ {9 x^{2} y^{5}={\color {Red} 3 x^{2} y^{3}} \cdot 3 y}\end{array} \nonumber \]
Observe that the set of second monomial factors (\(2x\) and \(3y\)) contain no additional common factors.
Find the greatest common factor of \(16xy^3\) and \(12x^4y^2\).
- Answer
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\(4 x y^{2}\)
Find the greatest common factor of \(12x^4\), \(18 x^3\), and \(30 x^2\).
Solution
To find the \(\mathrm{GCF}\) of \(12x^4\), \(18 x^3\), and \(30 x^2\), we note that:
- The greatest common factor of \(12\), \(18\), and \(30\) is \(6\).
- The monomials \(12x^4\), \(18 x^3\), and \(30 x^2\) have the variable \(x\) in common.
- The highest power of \(x\) in common is \(x^2\).
Thus, the greatest common factor is \(\mathrm{GCF}\left(12 x^{4}, 18 x^{3}, 30 x^{2}\right)=6 x^{2}\). Note what happens when we write each of the given monomials as a product of the greatest common factor and a second monomial:
\[\begin{array}{l}{12 x^{4}={\color {Red}6 x^{2}} \cdot 2 x^{2}} \\ {18 x^{3}={\color {Red}6 x^{2}} \cdot 3 x} \\ {30 x^{2}={\color {Red}6 x^{2}} \cdot 5}\end{array} \nonumber \]
Observe that the set of second monomial factors (\(2x^2\), \(3 x\), and \(5\)) contain no additional common factors.
Find the greatest common factor of \(6y^3\), \(15 y^2\), and \(9 y^5\).
- Answer
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\(3 y^{2}\)
Factor Out the GCF
Earlier, we multiplied a monomial and polynomial by distributing the monomial times each term in the polynomial.
\[\begin{aligned} {\color {Red}2 x(}3 x^{2}+4 x-7{\color {Red})} &={\color {Red}2 x} \cdot 3 x^{2}+{\color {Red}2 x} \cdot 4 x-{\color {Red}2 x} \cdot 7 \\ &=6 x^{3}+8 x^{2}-14 x \end{aligned} \nonumber \]
In this section we reverse that multiplication process using the distributive property, \[a \cdot b+a\cdot c=a(b+c)\nonumber\]
We present you with the final product and ask you to bring back the original multiplication problem. In the case \(6 x^{3}+8 x^{2}-14 x\), the greatest common factor of \(6x^3\), \(8x^2\), and \(14x\) is \(2x\). We then use the distributive property to factor out \(2x\) from each term of the polynomial.
\[\begin{aligned} 6 x^{3}+8 x^{2}-14 x &={\color {Red}2 x} \cdot 3 x^{2}+{\color {Red}2 x} \cdot 4 x-{\color {Red}2 x} \cdot 7 \\ &={\color {Red}2 x(}3 x^{2}+4 x-7{\color {Red})} \end{aligned} \nonumber \]
Factoring is the process of writing an expression as a product of polynomials.
Let’s look at a few examples that factor out the \(\mathrm{GCF}\).
Factor: \(6 x^{2}+10 x+14\)
Solution
The greatest common factor (\(\mathrm{GCF}\)) of \(6x^2\), \(10 x\) and \(14\) is \(2\). Factor out the \(\mathrm{GCF}\).
\[\begin{aligned} 6 x^{2}+10x+14 x &={\color {Red}2 } \cdot 3 x^{2}+{\color {Red}2 } \cdot 5 x+{\color {Red}2 } \cdot 7 \\ &={\color {Red}2 (}3 x^{2}+5 x+7{\color {Red})} \end{aligned} \nonumber \]
The factored form of \(6 x^{2}+10 x+14\) is \(2(3x^2+5x+7)\).
Check:
Check that you factored correctly by multiplying:
\[\begin{aligned} {\color {Red}2 (}3 x^{2}+5 x+7{\color {Red})} &={\color {Red}2 } \cdot 3 x^{2}+{\color {Red}2 } \cdot 5 x+{\color {Red}2 } \cdot 7 \\ &=6 x^{2}+10 x+14 \end{aligned} \nonumber \]
That’s the original polynomial, so we factored correctly.
Factor: \(9 y^{2}-15 y+12\)
- Answer
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\(3\left(3 y^{2}-5 y+4\right)\)
Factor: \(12 y^{5}-32 y^{4}+8 y^{2}\)
Solution
The greatest common factor (\(\mathrm{GCF}\)) of \(12y^5\), \(32y^4\) and \(8y^2\) is \(4y^2\). Factor out the \(\mathrm{GCF}\).
\[\begin{aligned} 12 y^{5}-32 y^{4}+8 y^{2} &={\color {Red}4 y^{2}} \cdot 3 y^{3}-{\color {Red}4 y^{2}} \cdot 8 y^{2}+{\color {Red}4 y^{2}} \cdot 2 \\ &={\color {Red}4 y^{2} (}3 y^{3}-8 y^{2}+2{\color {Red})} \end{aligned} \nonumber \]
Check: Multiply. Distribute the monomial \(4y^2\).
\[\begin{aligned} {\color {Red}4 y^{2} (}3 y^{3}-8 y^{2}+2{\color {Red})} &={\color {Red}4 y^{2}} \cdot 3 y^{3}-{\color {Red}4 y^{2}} \cdot 8 y^{2}+{\color {Red}4 y^{2}} \cdot 2 \\ &=12 y^{5}-32 y^{4}+8 y^{2} \end{aligned} \nonumber \]
That’s the original polynomial. We have factored correctly.
Factor: \(8 x^{6}+20 x^{4}-24 x^{3}\)
- Answer
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\(4 x^{3}\left(2 x^{3}+5 x-6\right)\)
Factor: \(12 a^{3} b+24 a^{2} b^{2}+12 a b^{3}\)
Solution
The greatest common factor (\(\mathrm{GCF}\)) of \(12a^3b\), \(24 a^2b^2\) and \(12ab^3\) is \(12ab\). Factor out the \(\mathrm{GCF}\).
\[\begin{aligned} 12 a^{3} b+24 a^{2} b^{2}+12 a b^{3} &={\color {Red}12ab} \cdot a^{2}-{\color {Red}12ab} \cdot 2 a b+{\color {Red}12ab} \cdot b^{2} \\ &= {\color {Red}12ab (}a^{2}+2ab+b^{2}{\color {Red})} \end{aligned} \nonumber \]
Check: Multiply. Distribute the monomial \(12ab\).
\[\begin{aligned} {\color {Red}12ab (}a^{2}+2ab+b^{2}{\color {Red})} &={\color {Red}12ab} \cdot a^{2}-{\color {Red}12ab} \cdot 2 a b+{\color {Red}12ab} \cdot b^{2} \\ &=12 a^{3} b+24 a^{2} b^{2}+12 a b^{3} \end{aligned} \nonumber \]
That’s the original polynomial. We have factored correctly.
Factor: \(15 s^{2} t^{4}+6 s^{3} t^{2}+9 s^{2} t^{2}\)
- Answer
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\(3 s^{2} t^{2}\left(5 t^{2}+2 s+3\right)\)
Remember that the distributive property allows us to pull the \(\mathrm{GCF}\) out in front of the expression or to pull it out in back. In symbols:
\({\color {Red}a}b+{\color {Red}a}c={\color {Red}a}(b+c) \quad\) or \(\quad b{\color {Red}a}+c{\color {Red}a}=(b+c){\color {Red}a}\)
We can also use the distributive property to factor out binomial factors.
Factor: \(2 x(3 x+2)+5(3 x+2)\)
Solution
In this case, the greatest common factor (\(\mathrm{GCF}\)) is \(3x + 2\).
\[\begin{aligned} 2x(3 x+2)+5(3 x+2) &=2x \cdot {\color {Red}(3 x+2)}+5 \cdot{\color {Red}(3 x+2)} \\ &=(2 x+5){\color {Red}(3 x+2)} \end{aligned} \nonumber \]
Because of the commutative property of multiplication, it is equally valid to pull the \(\mathrm{GCF}\) out in front.
\[\begin{aligned} 2x(3 x+2)+5(3 x+2) &=2x \cdot {\color {Red}(3 x+2)}+5 \cdot{\color {Red}(3 x+2)} \\ &={\color {Red}(3 x+2)} (2 x+5)\end{aligned}\]
Note that the order of factors differs from the first solution, but because of the commutative property of multiplication, the order does not matter. The answers are the same.
Factor: \(3 x^{2}(4 x-7)+8(4 x-7)\)
- Answer
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\(\left(3 x^{2}+8\right)(4 x-7)\)
Factor: \(15 a(a+b)-12(a+b)\)
Solution
In this case, the greatest common factor (\(\mathrm{GCF}\)) is \(3(a + b)\).
\[\begin{aligned} 15 a(a+b)-12(a+b) &={\color {Red}3(a+b)} \cdot 5 a-{\color {Red}3(a+b)} \cdot 4 \\ &={\color {Red}3(a+b)}(5 a-4) \end{aligned} \nonumber \]
Alternate solution:
It is possible that you might not notice that \(15\) and \(12\) are divisible by \(3\), factoring out only the common factor \(a + b\).
\[\begin{aligned} 15 a(a+b)-12(a+b) &=15a\cdot {\color {Red}(a+b)} - 12\cdot {\color {Red}(a+b)} \\ &=(15a-12) {\color {Red}(a+b)}\end{aligned} \nonumber \]
However, you now need to notice that you can continue to factor by factoring out \(3\) from both \(15a\) and \(12\).
\(=3(5 a-4)(a+b)\)
Note that the order of factors differs from the first solution, but because of the commutative property of multiplication, the order does not matter. The answers are the same.
Factor: \(24 m(m-2 n)+20(m-2 n)\)
- Answer
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\(4(6 m+5)(m-2 n)\)
Factoring by Grouping
The last factoring skill in this section involves four-term expressions. The technique for factoring a four-term expression is called factoring by grouping .
Factor by grouping: \(x^{2}+8 x+3 x+24\)
Solution
We “group” the first and second terms, noting that we can factor an \(x\) out of both of these terms. Then we “group” the third and fourth terms, noting that we can factor \(3\) out of both of these terms.
\[ \underbrace{x^2+8x} +\underbrace{3 x+24}=x(x+8)+3(x+8) \]
Now we can factor \(x + 8\) out of both of these terms.
\((x+3){\color {Red}(x+8)}\)
Factor by grouping: \(x^{2}-6 x+2 x-12\)
- Answer
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\((x+2)(x-6)\)
Let’s try a grouping that contains some negative signs.
Factor by grouping: \(x^{2}+4 x-7 x-28\)
Solution
We “group” the first and second terms, noting that we can factor \(x\) out of both of these terms. Then we “group” the third and fourth terms, then try to factor a \(7\) out of both these terms.
\[ \underbrace{x^2+4x} -\underbrace{7x-28}=x{\color{Red} (x+4)}+7{\color{Red} (-x-4)} \]
This does not lead to a common factor. Let’s try again, this time factoring a \(−7\) out of the third and fourth terms.
\[ \underbrace{x^2+4x} -\underbrace{7x-28}=x{\color{Red} (x+4)}-7{\color{Red} (x+4)} \]
That worked! We now factor out a common factor \(x + 4\).
\((x-7){\color {Red}(x+4)}\)
It is important to be careful with negatives when factoring by grouping. In the previous example, it would be incorrect to write \(x^{2}+4 x-7 x-28\) as \((x^2+4x)-(7x-28)\) because if you distribute the negative through, the last term is +28 which is not the original polynomial.
Factor by grouping: \(x^{2}-5 x-4 x+20\)
- Answer
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\((x-4)(x-5)\)
Let’s increase the size of the numbers a bit.
Factor by grouping: \(6 x^{2}-8 x+9 x-12\)
Solution
Note that we can factor \(2x\) out of the first two terms and \(3\) out of the second two terms.
\[ \underbrace{x^2+4x} -\underbrace{9x-12}=2x{\color{Red} (3 x-4)}-7{\color{Red} (3 x-4)} \]
Now we have a common factor \(3x−4\) which we can factor out.
\((2x+3){\color {Red}(3x-4)}\)
Factor by grouping: \(15 x^{2}+9 x+10 x+6\)
- Answer
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\((3 x+2)(5 x+3)\)
As the numbers get larger and larger, you need to factor out the (\(\mathrm{GCF}\)) from each grouping. If not, you won’t get a common factor to finish the factoring.
Factor by grouping: \(24 x^{2}-32 x-45 x+60\)
Solution
Suppose that we factor \(8x\) out of the first two terms and \(−5\) out of the second two terms.
\[ \underbrace{24x^2-32x} -\underbrace{45x+60}=8x{\color{Red} (3x-4)}-5{\color{Red} (9x-12)} \]
That did not work, as we don’t have a common factor to complete the factoring process. However, note that we can still factor out a \(3\) from \(9x−12\). As we’ve already factored out a \(5\), and now we see can factor out an additional \(3\), this means that we should have factored out \(3\) times \(5\), or \(15\), to begin with. Let’s start again, only this time we’ll factor \(15\) out of the second two terms.
\[ \underbrace{24x^2-32x} -\underbrace{45x+60}=8x{\color{Red} (3x-4)}-15{\color{Red} (3x-4)} \]
Beautiful! We can now factor out \(3x−4\).
\((8x-15){\color {Red}(3x-4)}\)
Factor by grouping: \(36 x^{2}-84 x+15 x-35\)
- Answer
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\((12 x+5)(3 x-7)\)