Learning Objectives
In this section, you will:
- Factor the greatest common factor of a polynomial.
- Factor a trinomial.
- Factor by grouping.
- Factor a perfect square trinomial.
- Factor a difference of squares.
- Factor the sum and difference of cubes.
- Factor expressions using fractional or negative exponents.
Imagine that we are trying to find the area of a lawn so that we can determine how much grass seed to purchase. The lawn is the green portion in Figure 1.
Figure 1
The area of the entire region can be found using the formula for the area of a rectangle.
A===lw10xβ 6x60x2units2π΄=ππ€=10π₯β 6π₯=60π₯2units2
The areas of the portions that do not require grass seed need to be subtracted from the area of the entire region. The two square regions each have an area of A=s2=42=16π΄=π 2=42=16 units2. The other rectangular region has one side of length 10xβ810π₯β8 and one side of length 4,4, giving an area of A=lw=4(10xβ8)=40xβ32π΄=ππ€=4(10π₯β8)=40π₯β32 units2. So the region that must be subtracted has an area of 2(16)+40xβ32=40×2(16)+40π₯β32=40π₯ units2.
The area of the region that requires grass seed is found by subtracting 60×2β40×60π₯2β40π₯ units2. This area can also be expressed in factored form as 20x(3xβ2)20π₯(3π₯β2) units2. We can confirm that this is an equivalent expression by multiplying.
Many polynomial expressions can be written in simpler forms by factoring. In this section, we will look at a variety of methods that can be used to factor polynomial expressions.
Factoring the Greatest Common Factor of a Polynomial
When we study fractions, we learn that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, 44 is the GCF of 1616 and 2020 because it is the largest number that divides evenly into both 1616 and 2020 The GCF of polynomials works the same way: 4×4π₯ is the GCF of 16×16π₯ and 20×220π₯2 because it is the largest polynomial that divides evenly into both 16×16π₯ and 20×2.20π₯2.
When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables.
GREATEST COMMON FACTOR
The greatest common factor (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials.
HOW TO
Given a polynomial expression, factor out the greatest common factor.
- Identify the GCF of the coefficients.
- Identify the GCF of the variables.
- Combine to find the GCF of the expression.
- Determine what the GCF needs to be multiplied by to obtain each term in the expression.
- Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.
EXAMPLE 1
Factoring the Greatest Common Factor
FactorΒ 6x3y3+45x2y2+21xy.6π₯3π¦3+45π₯2π¦2+21π₯π¦.
Analysis
After factoring, we can check our work by multiplying. Use the distributive property to confirm that (3xy)(2x2y2+15xy+7)=6x3y3+45x2y2+21xy.(3π₯π¦)(2π₯2π¦2+15π₯π¦+7)=6π₯3π¦3+45π₯2π¦2+21π₯π¦.
TRY IT #1
Factor x(b2βa)+6(b2βa)π₯(π2βπ)+6(π2βπ) by pulling out the GCF.
Factoring a Trinomial with Leading Coefficient 1
Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial x2+5x+6π₯2+5π₯+6 has a GCF of 1, but it can be written as the product of the factors (x+2)(π₯+2) and (x+3).(π₯+3).
Trinomials of the form x2+bx+cπ₯2+ππ₯+π can be factored by finding two numbers with a product of cπ and a sum of b.π. The trinomial x2+10x+16,π₯2+10π₯+16, for example, can be factored using the numbers 22 and 88 because the product of those numbers is 1616 and their sum is 10.10. The trinomial can be rewritten as the product of (x+2)(π₯+2) and (x+8).(π₯+8).
FACTORING A TRINOMIAL WITH LEADING COEFFICIENT 1
A trinomial of the form x2+bx+cπ₯2+ππ₯+π can be written in factored form as (x+p)(x+q)(π₯+π)(π₯+π) where pq=cππ=π and p+q=b.π+π=π.
Q&A
Can every trinomial be factored as a product of binomials?
No. Some polynomials cannot be factored. These polynomials are said to be prime.
HOW TO
Given a trinomial in the form x2+bx+c,π₯2+ππ₯+π, factor it.
- List factors ofΒ c.π.
- FindΒ pπΒ andΒ q,π,Β a pair of factors ofΒ cπΒ with a sum ofΒ b.π.
- Write the factored expressionΒ (x+p)(x+q).(π₯+π)(π₯+π).
EXAMPLE 2
Factoring a Trinomial with Leading Coefficient 1
FactorΒ x2+2xβ15.π₯2+2π₯β15.
Analysis
We can check our work by multiplying. Use FOIL to confirm that (xβ3)(x+5)=x2+2xβ15.(π₯β3)(π₯+5)=π₯2+2π₯β15.
Q&A
Does the order of the factors matter?
No. Multiplication is commutative, so the order of the factors does not matter.
TRY IT #2
Factor x2β7x+6.π₯2β7π₯+6.
Factoring by Grouping
Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial 2×2+5x+32π₯2+5π₯+3 can be rewritten as (2x+3)(x+1)(2π₯+3)(π₯+1) using this process. We begin by rewriting the original expression as 2×2+2x+3x+32π₯2+2π₯+3π₯+3 and then factor each portion of the expression to obtain 2x(x+1)+3(x+1).2π₯(π₯+1)+3(π₯+1). We then pull out the GCF of (x+1)(π₯+1) to find the factored expression.
FACTOR BY GROUPING
To factor a trinomial in the form ax2+bx+cππ₯2+ππ₯+π by grouping, we find two numbers with a product of acππ and a sum of b.π. We use these numbers to divide the xπ₯ term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression.
HOW TO
Given a trinomial in the form ax2+bx+c,ππ₯2+ππ₯+π, factor by grouping.
- List factors ofΒ ac.ππ.
- FindΒ pπΒ andΒ q,π,Β a pair of factors ofΒ acππΒ with a sum ofΒ b.π.
- Rewrite the original expression asΒ ax2+px+qx+c.ππ₯2+ππ₯+ππ₯+π.
- Pull out the GCF ofΒ ax2+px.ππ₯2+ππ₯.
- Pull out the GCF ofΒ qx+c.ππ₯+π.
- Factor out the GCF of the expression.
EXAMPLE 3
Factoring a Trinomial by Grouping
FactorΒ 5×2+7xβ65π₯2+7π₯β6Β by grouping.
Analysis
We can check our work by multiplying. Use FOIL to confirm that (5xβ3)(x+2)=5×2+7xβ6.(5π₯β3)(π₯+2)=5π₯2+7π₯β6.
TRY IT #3
Factor
- βΒ 2×2+9x+92π₯2+9π₯+9
- βΒ 6×2+xβ16π₯2+π₯β1
Factoring a Perfect Square Trinomial
A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.
a2+2ab+b2a2β2ab+b2=and=(a+b)2(aβb)2π2+2ππ+π2=(π+π)2andπ2β2ππ+π2=(πβπ)2
We can use this equation to factor any perfect square trinomial.
PERFECT SQUARE TRINOMIALS
A perfect square trinomial can be written as the square of a binomial:
a2+2ab+b2=(a+b)2π2+2ππ+π2=(π+π)2
HOW TO
Given a perfect square trinomial, factor it into the square of a binomial.
- Confirm that the first and last term are perfect squares.
- Confirm that the middle term is twice the product ofΒ ab.ππ.
- Write the factored form asΒ (a+b)2.(π+π)2.
EXAMPLE 4
Factoring a Perfect Square Trinomial
FactorΒ 25×2+20x+4.25π₯2+20π₯+4.
TRY IT #4
Factor 49×2β14x+1.49π₯2β14π₯+1.
Factoring a Difference of Squares
A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.
a2βb2=(a+b)(aβb)π2βπ2=(π+π)(πβπ)
We can use this equation to factor any differences of squares.
DIFFERENCES OF SQUARES
A difference of squares can be rewritten as two factors containing the same terms but opposite signs.
a2βb2=(a+b)(aβb)π2βπ2=(π+π)(πβπ)
HOW TO
Given a difference of squares, factor it into binomials.
- Confirm that the first and last term are perfect squares.
- Write the factored form asΒ (a+b)(aβb).(π+π)(πβπ).
EXAMPLE 5
Factoring a Difference of Squares
FactorΒ 9×2β25.9π₯2β25.
TRY IT #5
Factor 81y2β100.81π¦2β100.
Q&A
Is there a formula to factor the sum of squares?
No. A sum of squares cannot be factored.
Factoring the Sum and Difference of Cubes
Now, we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.
a3+b3=(a+b)(a2βab+b2)π3+π3=(π+π)(π2βππ+π2)
Similarly, the difference of cubes can be factored into a binomial and a trinomial, but with different signs.
a3βb3=(aβb)(a2+ab+b2)π3βπ3=(πβπ)(π2+ππ+π2)
We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: Same Opposite Always Positive. For example, consider the following example.
x3β23=(xβ2)(x2+2x+4)π₯3β23=(π₯β2)(π₯2+2π₯+4)
The sign of the first 2 is the same as the sign between x3β23.π₯3β23. The sign of the 2×2π₯ term is opposite the sign between x3β23.π₯3β23. And the sign of the last term, 4, is always positive.
SUM AND DIFFERENCE OF CUBES
We can factor the sum of two cubes as
a3+b3=(a+b)(a2βab+b2)π3+π3=(π+π)(π2βππ+π2)
We can factor the difference of two cubes as
a3βb3=(aβb)(a2+ab+b2)π3βπ3=(πβπ)(π2+ππ+π2)
HOW TO
Given a sum of cubes or difference of cubes, factor it.
- Confirm that the first and last term are cubes,Β a3+b3π3+π3Β orΒ a3βb3.π3βπ3.
- For a sum of cubes, write the factored form asΒ (a+b)(a2βab+b2).(π+π)(π2βππ+π2).Β For a difference of cubes, write the factored form asΒ (aβb)(a2+ab+b2).(πβπ)(π2+ππ+π2).
EXAMPLE 6
Factoring a Sum of Cubes
Factor x3+512.π₯3+512.
Analysis
After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.
TRY IT #6
Factor the sum of cubes: 216a3+b3.216π3+π3.
EXAMPLE 7
Factoring a Difference of Cubes
FactorΒ 8×3β125.8π₯3β125.
Analysis
Just as with the sum of cubes, we will not be able to further factor the trinomial portion.
TRY IT #7
Factor the difference of cubes: 1,000×3β1.1,000π₯3β1.
Factoring Expressions with Fractional or Negative Exponents
Expressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest power. These expressions follow the same factoring rules as those with integer exponents. For instance, 2×14+5×342π₯14+5π₯34 can be factored by pulling out x14π₯14 and being rewritten as x14(2+5×12).π₯14(2+5π₯12).
EXAMPLE 8
Factoring an Expression with Fractional or Negative Exponents
FactorΒ 3x(x+2)β13+4(x+2)23.3π₯(π₯+2)β13+4(π₯+2)23.
TRY IT #8
Factor 2(5aβ1)34+7a(5aβ1)β14.2(5πβ1)34+7π(5πβ1)β14.
MEDIA
Access these online resources for additional instruction and practice with factoring polynomials.
- Identify GCF
- Factor Trinomials when a Equals 1
- Factor Trinomials when a is not equal to 1
- Factor Sum or Difference of Cubes
1.5 Section Exercises
Verbal
1.
If the terms of a polynomial do not have a GCF, does that mean it is not factorable? Explain.
2.
A polynomial is factorable, but it is not a perfect square trinomial or a difference of two squares. Can you factor the polynomial without finding the GCF?
3.
How do you factor by grouping?
Algebraic
For the following exercises, find the greatest common factor.
4.
14x+4xyβ18xy214π₯+4π₯π¦β18π₯π¦2
5.
49mb2β35m2ba+77ma249ππ2β35π2ππ+77ππ2
6.
30x3yβ45x2y2+135xy330π₯3π¦β45π₯2π¦2+135π₯π¦3
7.
200p3m3β30p2m3+40m3200π3π3β30π2π3+40π3
8.
36j4k2β18j3k3+54j2k436π4π2β18π3π3+54π2π4
9.
6y4β2y3+3y2βy6π¦4β2π¦3+3π¦2βπ¦
For the following exercises, factor by grouping.
10.
6×2+5xβ46π₯2+5π₯β4
11.
2a2+9aβ182π2+9πβ18
12.
6c2+41c+636π2+41π+63
13.
6n2β19nβ116π2β19πβ11
14.
20w2β47w+2420π€2β47π€+24
15.
2p2β5pβ72π2β5πβ7
For the following exercises, factor the polynomial.
16.
7×2+48xβ77π₯2+48π₯β7
17.
10h2β9hβ910β2β9ββ9
18.
2b2β25bβ2472π2β25πβ247
19.
9d2β73d+89π2β73π+8
20.
90v2β181v+9090π£2β181π£+90
21.
12t2+tβ1312π‘2+π‘β13
22.
2n2βnβ152π2βπβ15
23.
16×2β10016π₯2β100
24.
25y2β19625π¦2β196
25.
121p2β169121π2β169
26.
4m2β94π2β9
27.
361d2β81361π2β81
28.
324×2β121324π₯2β121
29.
144b2β25c2144π2β25π2
30.
16a2β8a+116π2β8π+1
31.
49n2+168n+14449π2+168π+144
32.
121×2β88x+16121π₯2β88π₯+16
33.
225y2+120y+16225π¦2+120π¦+16
34.
m2β20m+100π2β20π+100
35.
25p2β120p+14425π2β120π+144
36.
36q2+60q+2536π2+60π+25
For the following exercises, factor the polynomials.
37.
x3+216π₯3+216
38.
27y3β827π¦3β8
39.
125a3+343125π3+343
40.
b3β8d3π3β8π3
41.
64×3β12564π₯3β125
42.
729q3+1331729π3+1331
43.
125r3+1,728s3125π3+1,728π 3
44.
4x(xβ1)β23+3(xβ1)134π₯(π₯β1)β23+3(π₯β1)13
45.
3c(2c+3)β14β5(2c+3)343π(2π+3)β14β5(2π+3)34
46.
3t(10t+3)13+7(10t+3)433π‘(10π‘+3)13+7(10π‘+3)43
47.
14x(x+2)β25+5(x+2)3514π₯(π₯+2)β25+5(π₯+2)35
48.
9y(3yβ13)15β2(3yβ13)659π¦(3π¦β13)15β2(3π¦β13)65
49.
5z(2zβ9)β32+11(2zβ9)β125π§(2π§β9)β32+11(2π§β9)β12
50.
6d(2d+3)β16+5(2d+3)566π(2π+3)β16+5(2π+3)56
Real-World Applications
For the following exercises, consider this scenario:
Charlotte has appointed a chairperson to lead a city beautification project. The first act is to install statues and fountains in one of the cityβs parks. The park is a rectangle with an area of 98×2+105xβ2798π₯2+105π₯β27 m2, as shown in the figure below. The length and width of the park are perfect factors of the area.
51.
Factor by grouping to find the length and width of the park.
52.
A statue is to be placed in the center of the park. The area of the base of the statue is 4×2+12x+9 m2.4π₯2+12π₯+9 m2. Factor the area to find the lengths of the sides of the statue.
53.
At the northwest corner of the park, the city is going to install a fountain. The area of the base of the fountain is 9×2β25 m2.9π₯2β25 m2. Factor the area to find the lengths of the sides of the fountain.
For the following exercise, consider the following scenario:
A school is installing a flagpole in the central plaza. The plaza is a square with side length 100 yd. as shown in the figure below. The flagpole will take up a square plot with area x2β6x+9π₯2β6π₯+9 yd2.
54.
Find the length of the base of the flagpole by factoring.
Extensions
For the following exercises, factor the polynomials completely.
55.
16×4β200×2+62516π₯4β200π₯2+625
56.
81y4β25681π¦4β256
57.
16z4β2,401a416π§4β2,401π4
58.
5x(3x+2)β24+(12x+8)325π₯(3π₯+2)β24+(12π₯+8)32
59.
(32×3+48×2β162xβ243)β1
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