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
4x2(2x2−3x+8)=8x4−12x3+32x2
We work out the same problem, but backwards. We will start with 8x2−12x3+32x2 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:
15=3⋅524=23⋅327=33
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 31 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, 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 24x4y2z, 18x2y4, and 12x3yz5
Solution
First we obtain the prime factorization of each monomial:
24x4y2z=23⋅3⋅x4⋅y2⋅z18x2y4=2⋅32⋅x2⋅y412x3yz5=22⋅3⋅x3⋅y⋅z5
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⋅3⋅x2⋅y⋅z. Hence, GCF(24x4y2z,18x2y4,12x3yz5)=6x2yz.
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: 4x2−20x+16
Solution
Looking at each term, let’s write the prime factorization of each term:
4x2=22⋅x220x=22⋅5⋅x16=24
We need to take 22. Hence, GCF(4x2,20x,16)=22=4. Let’s rewrite each term in the expression as the product of the GCF and the factors left:
4x2−20x+16Rewrite with the GCF 44⋅x2−4⋅5x+4⋅4Rewrite the expression with the GCF and parenthesis4(x2−5x+4)Factored form
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. ✓ Verify the factored form by multiplying. The product should be the original expression.
Factor out the GCF: 25x4−15x3+20x2
Solution
Looking at each term, let’s start by writing the prime factorization of each term.
25x4=52⋅x415x3=3⋅5⋅x320x2=22⋅5⋅x2
Step 1. We need to take 5x2. Hence, GCF(25x4,15x3,20x2)=5x2.
Step 2. Let’s rewrite each term in the expression as the product of the GCF and the factors left: 25x4−15x3+20x2Rewrite with the GCF 5x25x2⋅5x2−5x2⋅3x+5x2⋅4
Step 3. Rewrite the expression with the GCF and the remaining factors in parenthesis: 5x2⋅5x2−5x2⋅3x+5x2⋅4Rewrite the expression with the GCF and parenthesis5x2(5x2−3x+4)Factored form
Step 4. Let’s verify the factored form: 5x2(5x2−3x+4)Distribute the GCF25x4−15x3+20x2✓ Original expression
Thus, the factored form is 5x2(5x2−3x+4).
Factor out the GCF: 3x3y2z+5x4y3z5−4xy4
Solution
Looking at each term, let’s start by writing the prime factorization of each term.
3x3y2z=3⋅x3⋅y2⋅z5x4y3z5=5⋅x4⋅y3⋅z54xy4=22⋅x⋅y4
Step 1. We need to take xy2. Hence, GCF(3x3y2z,5x4y3z5,4xy4)=xy2.
Step 2. Let’s rewrite each term in the expression as the product of the GCF and the factors left: 3x3y2z+5x4y3z5−4xy4Rewrite with the GCF xy2xy2⋅3x2z+xy2⋅5x3yz5−xy2⋅4y2
Step 3. Rewrite the expression with the GCF and the remaining factors in parenthesis: xy2⋅3x2z+xy2⋅5x3yz5−xy2⋅4y2Rewrite the expression with the GCF and parenthesisxy2(3x2z+5x2yz5−4y2)Factored form
Step 4. Let’s verify the factored form: xy2(3x2z+5x2yz5−4y2)Distribute the GCF3x3y2z+5x4y3z5−4xy4✓ Original expression
Thus, the factored form is xy2(3x2z+5x2yz5−4y2).
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: 21x3+14x2+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.
21x3+14x2+7xRewrite with the GCF 7x7x⋅3x2+7x⋅2x+7x⋅1Rewrite the expression with the GCF and parenthesis7x(3x2+2x+1)Factored form
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
3ax−7bxBoth have x in common, factor it outx(3a−7b)Factored form
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
3a(2a+5b)−7b(2a+5b)Both have (2a+5b) in common, factor it out(2a+5b)(3a−7b)Factored form
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
(2a+3)(5b+2)Distribute (2a+3) into second parenthesis5b(2a+3)+2(2a+3)Distribute10ab+15b+4a+6Product
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: 10ab+15b+4a+6Group the first two terms and the last two terms(10ab+15b)+(4a+6)
Step 2. Factor the GCF from each group: (10ab+15b)+(4a+6)Factor 5b from the first group and 2 from the second group5b(2a+3)+2(2a+3)
Step 3. Factor the GCF from the expression: 5b(2a+3)+2(2a+3)Factor the GCF (2a+3)(2a+3)(5b+2)Factored form
Recall, we can verify the factored form by multiplying the binomials and obtaining the original expression.
Factor: 6x2+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: 6x2+9xy−14x−21yGroup the first two terms and the last two terms(6x2+9xy)+(−14x−21y)
Step 2. Factor the GCF from each group. (6x2+9xy)+(−14x−21y)Factor 3x from the first group and −7from the second group3x(2x+3y)−7(2x+3y)
Step 3. Factor the GCF from the expression: 3x(2x+3y)−7(2x+3y)Factor the GCF (2x+3y)(2x+3y)(3x−7)Factored form
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: 5xy−8x−10y+16Group the first two terms and the last two terms(5xy−8x)+(−10y+16)
Step 2. Factor the GCF from each group: (5xy−8x)+(−10y+16)Factor x from the first group and −2from the second groupx(5y−8)−2(5y−8)Both binomials are identical
Step 3. Factor the GCF from the expression: x(5y−8)−2(5y−8)Factor the GCF (5y−8)(5y−8)(x−2)Factored form
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: 12ab−14a−6b+7Group the first two terms and the last two terms(12ab−14a)+(−6b+7)
Step 2. Factor the GCF from each group: (12ab−14a)+(−6b+7)Factor 2a from the first group and −1from the second group2a(6b−7)−1(6b−7)Both binomials are identical
Step 3. Factor the GCF from the expression: 2a(6b−7)−1(6b−7)Factor the GCF (6b−7)(6b−7)(2a−1)Factored form
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: 4a2−21b3+6ab−14ab2
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: 4a2−21b3+6ab−14ab2Group the first two terms and the last two terms(4a2−21b3)+(6ab−14ab2)
Step 2. Factor the GCF from each group: (4a2−21b3)+(6ab−14ab2)Factor 2ab from the second group(4a2−21b3)+2a(3b−7b2)Binomials are NOT identical
Step 1. Group two sets of two terms: 4a2+6ab−21b3−14ab2Group the first two terms and the last two terms(4a2+6ab)+(−21b3−14ab2)
Step 2. Factor the GCF from each group: (4a2+6ab)+(−21b3−14ab2)Factor 2a from the first group and −7b2from the second group2a(2a+3b)−7b2(3b+2a)Rewrite so the binomials are identical2a(2a+3b)−7b2(2a+3b)Binomials are identical
Step 3. Factor the GCF from the expression: 2a(2a+3b)−7b2(2a+3b)Factor the GCF (2a+3b)(2a+3b)(2a−7b2)Factored form
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: 8xy−12y+15−10xGroup the first two terms and the last two terms(8xy−12y)+(15−10x)
Step 2. Factor the GCF from each group: (8xy−12y)+(15−10x)Factor 4y from the first group and 5from the second group4y(2x−3)+5(3−2x)Binomials are NOT identical, but VERY close
Step 3. Factor the GCF from the expression: 4y(2x−3)−5(2x−3)Factor the GCF (2x−3)(2x−3)(4y−5)Factored form
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.
(b−a)Factor out −1−1(−b+a)Apply the commutative property to the addition−1(a−b)Now the binomial is written as subtraction
Sofia Kovalevskaya of Russia was the first woman on the editorial staff of a mathematical journal in the late 19th century. She also researched the planet Saturn’s rotating rings.
Greatest Common Factor and Grouping Homework
Factor the greatest common factor.
9+8b2
45x2−25
56−35p
7ab−35a2b
−3a2b+6a3b2
−5x2−5x3−15x4
20x4−30x+30
28m4+40m3+8
30b9+5ab−15a2
−48a2b2−56a3b−56a5b
20x8y2z2+15x5y2z+35x3y3z
50x2y+10y2+70xz2
30qpr−5qp+5q
−18n5+3n3−21n+3
−40x11−20x12+50x13−50x14
−32mn8+4m6n+12mn4+16mn
x−5
1+2n2
50x−80y
27x2y5−72x3y2
8x3y2+4x3
−32n9+32n6+40n5
21p6+30p2+27
−10x4+20x2+12x
27y7+12y2x+9y2
30m6+12mn2−25
3p+12q−15q2r2
30y4z3x5+50y4z5−10y4z3x
28b+14b2+35b3+7b5
30a8+6a5+27a3+21a2
−24x6−4x4+12x3+4x2
−10y7+6y10−4y10x−8y8x
Factor each completely.
40r3−8r2−25r+5
3n3−2n2−9n+6
15b3+21b2−35b−49
3x3+15x2+2x+10
35x3−28x2−20x+16
7xy−49x+5y−35
32xy+40x2+12y+15x
16xy−56x+2y−7
2xy−8x2+7y3−28y2x
40xy+35x−8y2−7y
32uv−20u+24v−15
10xy+30+25x+12y
3uv+14u−6u2−7v
16xy−3x−6x2+8y
35x3−10x2−56x+16
14v3+10v2−7v−5
6x3−48x2+5x−40
28p3+21p2+20p+15
7n3+21n2−5n−15
42r3−49r2+18r−21
15ab−6a+5b3−2b2
3mn−8m+15n−40
5mn+2m−25n−10
8xy+56x−y−7
4uv+14u2+12v+42u
24xy+25y2−20x−30y3
56ab+14−49a−16b