7.1: Greatest common factor and grouping
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In this lesson, we focus on factoring using the greatest common factor, GCF, of a polynomial. When we multiplied polynomials, we multiplied monomials by polynomials by distributing, such as
\[4x^2(2x^2-3x+8)=8x^4-12x^3+32x^2\nonumber\]
We work out the same problem, but backwards. We will start with \(8x^2 −12x^3 + 32x^2\) and obtain its factored form.
First, we have to identify the GCF of a polynomial. We introduce the GCF of a polynomial by looking at an example in arithmetic. The method in which we obtained the GCF between numbers in arithmetic is the same method we use to obtain the GCF with polynomials.
Finding the Greatest Common Factor
The factored form of a number or expression is the expression written as a product of factors.
The greatest common factor (GCF) of a polynomial is the largest polynomial that is a factor of all terms in the polynomial.
Find the GCF of \(15\), \(24\), and \(27\).
Solution
First we obtain the prime factorization of each number:
\[\begin{aligned}15&=3\cdot 5 \\ 24&=2^3\cdot 3\\ 27&=3^3\end{aligned}\]
Next, we take only the common factors and if any common factors repeat, we take the factor with the smallest exponent. Recall, the GCF is the largest factor that divides into all numbers. We need to take \(3\) (we only take \(3^1\) because there is only one three in common in all three numbers). Notice, there are no other factors in common with all three numbers. Hence, \(\text{GCF}(15, 24, 27) = 3\).
When finding the GCF of a polynomial, we use the same method and the only difference is that we have variables in the expression.
Find the GCF of \(24x^4y^2z\), \(18x^2y^4\), and \(12x^3yz^5\)
Solution
First we obtain the prime factorization of each monomial:
\[\begin{aligned}24x^4y^2z&=2^3\cdot 3\cdot x^4\cdot y^2\cdot z \\ 18x^2y^4&=2\cdot 3^2\cdot x^2\cdot y^4 \\ 12x^3yz^5&=2^2\cdot 3\cdot x^3\cdot y\cdot z^5\end{aligned}\]
Next, we take only the common factors and if any common factors repeat, we take the factor with the smallest exponent. Recall, the GCF is the largest factor that divides into all terms in the expression. We need to take \(2\cdot 3\cdot x^2\cdot y\cdot z\). Hence, \(\text{GCF}(24x^4y^2z,\: 18x^2y^4 ,\: 12x^3yz^5) = 6x^2yz\).
Factoring the Greatest Common Factor
Once we obtain the GCF, we can start factoring the GCF from an expression. Eventually, we want to be able to see the GCF quickly and begin factoring right away. Of course, only with practice we obtain better skills. Let’s try an example.
Factor out the GCF: \(4x^2-20x+16\)
Solution
Looking at each term, let’s write the prime factorization of each term:
\[\begin{aligned}4x^2&=2^2\cdot x^2 \\ 20x&=2^2\cdot 5\cdot x \\ 16&=2^4\end{aligned}\]
We need to take \(2^2\). Hence, \(\text{GCF}(4x^2,\: 20x,\: 16) = 2^2 = 4\). Let’s rewrite each term in the expression as the product of the GCF and the factors left:
\[\begin{array}{rl}4x^2-20x+16&\text{Rewrite with the GCF }4 \\ \color{blue}{4}\color{black}{}\cdot x^2-\color{blue}{4}\color{black}{}\cdot 5x+\color{blue}{4}\color{black}{}\cdot 4&\text{Rewrite the expression with the GCF and parenthesis} \\ \color{blue}{4}\color{black}{}(x^2-5x+4)&\text{Factored form}\end{array}\nonumber\]
Step 1. Find the GCF of the expression.
Step 2. Rewrite each term as a product of the GCF and the remaining factors.
Step 3. Rewrite as a product of the GCF and the remaining factors in parenthesis.
Step 4. \(\checkmark\) Verify the factored form by multiplying. The product should be the original expression.
Factor out the GCF: \(25x^4-15x^3+20x^2\)
Solution
Looking at each term, let’s start by writing the prime factorization of each term.
\[\begin{aligned}25x^4&=5^2\cdot x^4 \\ 15x^3&=3\cdot 5\cdot x^3 \\ 20x^2&=2^2\cdot 5\cdot x^2\end{aligned}\]
Step 1. We need to take \(5x^2\). Hence, \(\text{GCF}(25x^4,\: 15x^3,\: 20x^2) = 5x^2\).
Step 2. Let’s rewrite each term in the expression as the product of the GCF and the factors left: \[\begin{array}{rl}25x^4-15x^3+20x^2&\text{Rewrite with the GCF }5x^2 \\ \color{blue}{5x^2}\color{black}{}\cdot 5x^2-\color{blue}{5x^2}\color{black}{}\cdot 3x+\color{blue}{5x^2}\color{black}{}\cdot 4\end{array}\nonumber\]
Step 3. Rewrite the expression with the GCF and the remaining factors in parenthesis: \[\begin{array}{rl}\color{blue}{5x^2}\color{black}{}\cdot 5x^2-\color{blue}{5x^2}\color{black}{}\cdot 3x+\color{blue}{5x^2}\color{black}{}\cdot 4&\text{Rewrite the expression with the GCF and parenthesis} \\ \color{blue}{5x^2}\color{black}{}(5x^2-3x+4)&\text{Factored form}\end{array}\nonumber\]
Step 4. Let’s verify the factored form: \[\begin{array}{rl}\color{blue}{5x^2}\color{black}{}(5x^2-3x+4)&\text{Distribute the GCF} \\ 25x^4-15x^3+20x^2&\checkmark\text{ Original expression}\end{array}\nonumber\]
Thus, the factored form is \(5x^2(5x^2-3x+4)\).
Factor out the GCF: \(3x^3y^2z + 5x^4y^3z^5 − 4xy^4\)
Solution
Looking at each term, let’s start by writing the prime factorization of each term.
\[\begin{aligned}3x^3y^2z&=3\cdot x^3\cdot y^2\cdot z \\ 5x^4y^3z^5&=5\cdot x^4\cdot y^3\cdot z^5 \\ 4xy^4&=2^2\cdot x\cdot y^4\end{aligned}\]
Step 1. We need to take \(xy^2\). Hence, \(\text{GCF}(3x^3y^2z,\: 5x^4y^3z^5,\: 4xy^4) = xy^2\).
Step 2. Let’s rewrite each term in the expression as the product of the GCF and the factors left: \[\begin{array}{rl}3x^3y^2z+5x^4y^3z^5-4xy^4&\text{Rewrite with the GCF }xy^2 \\ \color{blue}{xy^2}\color{black}{}\cdot 3x^2z+\color{blue}{xy^2}\color{black}{}\cdot 5x^3yz^5-\color{blue}{xy^2}\color{black}{}\cdot 4y^2\end{array}\nonumber\]
Step 3. Rewrite the expression with the GCF and the remaining factors in parenthesis: \[\begin{array}{rl}\color{blue}{xy^2}\color{black}{}\cdot 3x^2z+\color{blue}{xy^2}\color{black}{}\cdot 5x^3yz^5-\color{blue}{xy^2}\color{black}{}\cdot 4y^2&\text{Rewrite the expression with the GCF and parenthesis} \\ \color{blue}{xy^2}\color{black}{}(3x^2z+5x^2yz^5-4y^2)&\text{Factored form}\end{array}\nonumber\]
Step 4. Let’s verify the factored form: \[\begin{array}{rl}\color{blue}{xy^2}\color{black}{}(3x^2z+5x^2yz^5-4y^2)&\text{Distribute the GCF} \\ 3x^3y^2z+5x^4y^3z^5-4xy^4&\checkmark\text{ Original expression}\end{array}\nonumber\]
Thus, the factored form is \(xy^2 (3x^2z + 5x^2yz^5 − 4y^2 )\).
The first recorded algorithm for finding the greatest common factor comes from Greek mathematician Euclid around the year 300 BC.
Let’s try an example with eye-balling the GCF and rewriting in factored form.
Factor out the GCF: \(21x^3+14x^2+7x\)
Solution
Looking at the coefficients, we can see that there is common factor of \(7\) in each term. Furthermore, we see a factor of \(x\) in common in all three terms. Hence, we take \(7x\) as the GCF. Notice we didn’t take a larger exponent on \(x\) because only one factor of \(x\) is common in all three terms.
Let's rewrite the expression in factored form.
\[\begin{array}{rl}21x^3+14x^2+7x&\text{Rewrite with the GCF }7x \\ \color{blue}{7x}\color{black}{}\cdot 3x^2+\color{blue}{7x}\color{black}{}\cdot 2x+\color{blue}{7x}\color{black}{}\cdot 1 &\text{Rewrite the expression with the GCF and parenthesis} \\ \color{blue}{7x}\color{black}{}(3x^2+2x+1)&\text{Factored form}\end{array}\nonumber\]
We can always verify the factored form by distributing the \(7x\) and obtaining the original expression.
A Binomial as the Greatest Common Factor
As part of a general strategy for factoring, we always look for a GCF. Sometimes the GCF is a monomial, like in the previous examples, or a binomial. Here we discuss factoring a polynomial where the GCF is a binomial. We usually call this factor by grouping. Consider the below example.
Factor: \(3ax-7bx\)
Solution
\[\begin{array}{rl}3a\color{blue}{x}\color{black}{}-7b\color{blue}{x}&\text{Both have }x\text{ in common, factor it out} \\ \color{blue}{x}\color{black}{}(3a-7b)&\text{Factored form}\end{array}\nonumber\]
Let’s make this interesting. Let’s apply this same method, but instead of the GCF being \(x\), let the GCF be the binomial \(2a + 5b\).
Find: \(3a(2a+5b)-7b(2a+5b)\)
Solution
\[\begin{array}{rl}3a\color{blue}{(2a+5b)}\color{black}{}-7b\color{blue}{(2a+5b)}&\color{black}{\text{Both have }}(2a+5b)\text{ in common, factor it out} \\ \color{blue}{(2a+5b)}\color{black}{}(3a-7b)&\text{Factored form}\end{array}\nonumber\]
Factor by Grouping
In Example 7.1.8 , we factored out a GCF of \((2a + 5b)\) the same way we factored out an \(x\) in Example 7.1.7 . This process can be extended to factor expressions where there isn’t a GCF. We will use a process known as factor by grouping. Factor by grouping is a method used to factor polynomials when there is at least four terms in the expression. Take the next example.
Multiply: \((2a+3)(5b+2)\)
Solution
\[\begin{array}{rl}(2a+3)(5b+2)&\text{Distribute }(2a+3)\text{ into second parenthesis} \\ 5b(2a+3)+2(2a+3)&\text{Distribute} \\ 10ab+15b+4a+6&\text{Product}\end{array}\nonumber\]
Notice the product has four terms none of which share a common factor.
To factor by grouping, we first notice the polynomial expression obtains four terms.
Step 1. Group two sets of two terms, e.g., \(ax + ay + bx + by = (ax + ay) + (bx + by)\).
Step 2. Factor the GCF from each group, e.g., \(a(x + y) + b(x + y)\)
Step 3. Factor the GCF from the expression, e.g., \((x + y)(a + b)\).
Factor: \(10ab+15b+4a+6\)
Solution
Notice we have \(4\) terms none of which share a common factor. Hence, we use factor by grouping.
Step 1. Group two sets of two terms: \[\begin{array}{rl}10ab+15b+4a+6&\text{Group the first two terms and the last two terms} \\ (10ab+15b)+(4a+6)\end{array}\nonumber\]
Step 2. Factor the GCF from each group: \[\begin{array}{rl}(10ab+15b)+(4a+6)&\text{Factor }5b\text{ from the first group and }2\text{ from the second group} \\ 5b(2a+3)+2(2a+3)\end{array}\nonumber\]
Step 3. Factor the GCF from the expression: \[\begin{array}{rl}5b\color{blue}{(2a+3)}\color{black}{}+2\color{blue}{(2a+3)}&\color{black}{\text{Factor the GCF }}(2a+3) \\ \color{blue}{(2a+3)}\color{black}{}(5b+2)&\text{Factored form}\end{array}\nonumber\]
Recall, we can verify the factored form by multiplying the binomials and obtaining the original expression.
Factor: \(6x^2+9xy-14x-21y\)
Solution
Notice we have \(4\) terms none of which share a common factor. Hence, we use factor by grouping.
Step 1. Group two sets of two terms: \[\begin{array}{rl}6x^2+9xy-14x-21y&\text{Group the first two terms and the last two terms} \\ (6x^2+9xy)+(-14x-21y)\end{array}\nonumber\]
Step 2. Factor the GCF from each group. \[\begin{array}{rl}(6x^2+9xy)+(-14x-21y)&\text{Factor }3x\text{ from the first group and }-7 \\ &\text{from the second group} \\ 3x(2x+3y)-7(2x+3y)\end{array}\nonumber\]
Step 3. Factor the GCF from the expression: \[\begin{array}{rl}3x\color{blue}{(2x+3y)}\color{black}{}-7\color{blue}{(2x+3y)}&\color{black}{\text{Factor the GCF }}(2x+3y) \\ \color{blue}{(2x+3y)}\color{black}{}(3x-7)&\text{Factored form}\end{array}\nonumber\]
Recall, we can verify the factored form by multiplying the binomials and obtaining the original expression.
Notice after Step 2., we want the binomial’s GCFto be identical so that we can factor it out in Step 3. Be sure these binomials are identical. A common error is when the binomials aren’t identical (sometimes by a negative) and students factor anyways.
Factor: \(5xy-8x-10y+16\)
Solution
Notice we have \(4\) terms none of which share a common factor. Hence, we use factor by grouping.
Step 1. Group two sets of two terms: \[\begin{array}{rl}5xy-8x-10y+16&\text{Group the first two terms and the last two terms} \\ (5xy-8x)+(-10y+16)\end{array}\nonumber\]
Step 2. Factor the GCF from each group: \[\begin{array}{rl}(5xy-8x)+(-10y+16)&\text{Factor }x\text{ from the first group and }-2 \\ &\text{from the second group} \\ x(5y-8)-2(5y-8)&\text{Both binomials are identical}\end{array}\nonumber\]
Step 3. Factor the GCF from the expression: \[\begin{array}{rl}x\color{blue}{(5y-8)}\color{black}{}-2\color{blue}{(5y-8)}&\color{black}{\text{Factor the GCF }}(5y-8) \\ \color{blue}{(5y-8)}\color{black}{}(x-2)&\text{Factored form}\end{array}\nonumber\]
Recall, we can verify the factored form by multiplying the binomials and obtaining the original expression.
Factor: \(12ab-14a-6b+7\)
Solution
Notice we have \(4\) terms none of which share a common factor. Hence, we use factor by grouping.
Step 1. Group two sets of two terms: \[\begin{array}{rl}12ab-14a-6b+7&\text{Group the first two terms and the last two terms} \\ (12ab-14a)+(-6b+7)\end{array}\nonumber\]
Step 2. Factor the GCF from each group: \[\begin{array}{rl}(12ab-14a)+(-6b+7)&\text{Factor }2a\text{ from the first group and }-1 \\ &\text{from the second group} \\ 2a(6b-7)-1(6b-7)&\text{Both binomials are identical}\end{array}\nonumber\]
Step 3. Factor the GCF from the expression: \[\begin{array}{rl}2a\color{blue}{(6b-7)}\color{black}{}-1\color{blue}{(6b-7)}&\color{black}{\text{Factor the GCF }}(6b-7) \\ \color{blue}{(6b-7)}\color{black}{}(2a-1)&\text{Factored form}\end{array}\nonumber\]
Careful in these types of expressions, where we factor the entire second binomial and are left with the term \(1\). This occurs sometimes with factoring and it’s important to always write the \(1\) in Step 2. so that we do not forget it is there. Recall, we can verify the factored form by multiplying the binomials and obtaining the original expression.
Factor by Grouping by Rearranging Terms
Sometimes after completing Step 2., the binomials aren’t identical (by more than a negative sign). At this point we must return to the original problem and rearrange the terms so that when we factor by grouping, we obtain identical binomials in Step 2.
Factor: \(4a^2 − 21b^3 + 6ab − 14ab^2\)
Solution
Notice we have \(4\) terms none of which share a common factor. Hence, we use factor by grouping.
Step 1. Group two sets of two terms: \[\begin{array}{rl}4a^2-21b^3+6ab-14ab^2&\text{Group the first two terms and the last two terms} \\ (4a^2-21b^3)+(6ab-14ab^2)\end{array}\nonumber\]
Step 2. Factor the GCF from each group: \[\begin{array}{rl}(4a^2-21b^3)+(6ab-14ab^2)&\text{Factor }2ab\text{ from the second group} \\ (4a^2-21b^3)+2a(3b-7b^2)&\text{Binomials are NOT identical}\end{array}\nonumber\] Since these binomials aren’t identical, we return to the original expression and rearrange the terms. Let’s try moving \(6ab\) to the first group and \(−21b^3\) to the second group.
Step 1. Group two sets of two terms: \[\begin{array}{rl}4a^2+6ab-21b^3-14ab^2&\text{Group the first two terms and the last two terms} \\ (4a^2+6ab)+(-21b^3-14ab^2) \end{array}\nonumber\]
Step 2. Factor the GCF from each group: \[\begin{array}{rl}(4a^2+6ab)+(-21b^3-14ab^2)&\text{Factor }2a\text{ from the first group and }-7b^2 \\ &\text{from the second group} \\ 2a(2a+3b)-7b^2(3b+2a)&\text{Rewrite so the binomials are identical} \\ 2a(2a+3b)-7b^2(2a+3b)&\text{Binomials are identical}\end{array}\nonumber\]
Step 3. Factor the GCF from the expression: \[\begin{array}{rl}2a\color{blue}{(2a+3b)}\color{black}{}-7b^2\color{blue}{(2a+3b)}&\color{black}{\text{Factor the GCF }}(2a+3b) \\ \color{blue}{(2a+3b)}\color{black}{}(2a-7b^2)&\text{Factored form}\end{array}\nonumber\]
Recall, we can verify the factored form by multiplying the binomials and obtaining the original expression.
Factor: \(8xy-12y+15-10x\)
Solution
Notice we have \(4\) terms none of which share a common factor. Hence, we use factor by grouping.
Step 1. Group two sets of two terms: \[\begin{array}{rl}8xy-12y+15-10x&\text{Group the first two terms and the last two terms} \\ (8xy-12y)+(15-10x)\end{array}\nonumber\]
Step 2. Factor the GCF from each group: \[\begin{array}{rl}(8xy-12y)+(15-10x)&\text{Factor }4y\text{ from the first group and }5 \\ &\text{from the second group} \\ 4y(2x-3)+5(3-2x)&\text{Binomials are NOT identical, but VERY close}\end{array}\nonumber\] Since these binomials aren’t identical but close to it, we can think about it some more. These binomials would be identical if only the \(3\) and \(−2x\) in the second binomial were switched. Let’s factor a \(−1\) out of the second binomial: \[\begin{array}{rl}4y(2x-3)+5(3-2x)&\text{Factor a }-1\text{ from the second binomial} \\ 4y(2x-3)+5\cdot\color{blue}{-1}\color{black}{}(-3+2x)&\text{Rewrite so the binomials are identical} \\ 4y(2x-3)-5(2x-3)&\text{Binomials are identical}\end{array}\nonumber\]
Step 3. Factor the GCF from the expression: \[\begin{array}{rl}4y\color{blue}{(2x-3)}\color{black}{}-5\color{blue}{(2x-3)}&\color{black}{\text{Factor the GCF }}(2x-3) \\ \color{blue}{(2x-3)}\color{black}{}(4y-5)&\text{Factored form}\end{array}\nonumber\]
Recall, we can verify the factored form by multiplying the binomials and obtaining the original expression.
If the binomials are \((a−b)\) and \((b−a)\), we will factor \(−1\) out of one of the binomials to obtain identical binomials.
\[\begin{array}{rl}(b-a)&\text{Factor out }-1 \\ -1(-b+a)&\text{Apply the commutative property to the addition} \\ -1(a-b)&\text{Now the binomial is written as subtraction}\end{array}\nonumber\]
Sofia Kovalevskaya of Russia was the first woman on the editorial staff of a mathematical journal in the late \(19^{\text{th}}\) century. She also researched the planet Saturn’s rotating rings.
Greatest Common Factor and Grouping Homework
Factor the greatest common factor.
\(9+8b^2\)
\(45x^2-25\)
\(56-35p\)
\(7ab-35a^2b\)
\(-3a^2b+6a^3b^2\)
\(-5x^2-5x^3-15x^4\)
\(20x^4-30x+30\)
\(28m^4+40m^3+8\)
\(30b^9+5ab-15a^2\)
\(-48a^2b^2-56a^3b-56a^5b\)
\(20x^8y^2z^2+15x^5y^2z+35x^3y^3z\)
\(50x^2y+10y^2+70xz^2\)
\(30qpr − 5qp + 5q\)
\(-18n^5+3n^3-21n+3\)
\(-40x^{11}-20x^{12}+50x^{13}-50x^{14}\)
\(-32mn^8+4m^6n+12mn^4+16mn\)
\(x-5\)
\(1+2n^2\)
\(50x-80y\)
\(27x^2y^5-72x^3y^2\)
\(8x^3y^2+4x^3\)
\(-32n^9+32n^6+40n^5\)
\(21p^6+30p^2+27\)
\(-10x^4+20x^2+12x\)
\(27y^7+12y^2x+9y^2\)
\(30m^6+12mn^2-25\)
\(3p+12q-15q^2r^2\)
\(30y^4z^3x^5+50y^4z^5-10y^4z^3x\)
\(28b+14b^2+35b^3+7b^5\)
\(30a^8+6a^5+27a^3+21a^2\)
\(-24x^6-4x^4+12x^3+4x^2\)
\(-10y^7+6y^{10}-4y^{10}x-8y^8x\)
Factor each completely.
\(40r^3-8r^2-25r+5\)
\(3n^3-2n^2-9n+6\)
\(15b^3+21b^2-35b-49\)
\(3x^3+15x^2+2x+10\)
\(35x^3-28x^2-20x+16\)
\(7xy-49x+5y-35\)
\(32xy+40x^2+12y+15x\)
\(16xy-56x+2y-7\)
\(2xy-8x^2+7y^3-28y^2x\)
\(40xy+35x-8y^2-7y\)
\(32uv − 20u + 24v − 15\)
\(10xy + 30 + 25x + 12y\)
\(3uv+14u-6u^2-7v\)
\(16xy-3x-6x^2+8y\)
\(35x^3-10x^2-56x+16\)
\(14v^3+10v^2-7v-5\)
\(6x^3-48x^2+5x-40\)
\(28p^3+21p^2+20p+15\)
\(7n^3+21n^2-5n-15\)
\(42r^3-49r^2+18r-21\)
\(15ab-6a+5b^3-2b^2\)
\(3mn − 8m + 15n − 40\)
\(5mn + 2m − 25n − 10\)
\(8xy + 56x − y − 7\)
\(4uv+14u^2+12v+42u\)
\(24xy+25y^2-20x-30y^3\)
\(56ab + 14 − 49a − 16b\)