Chapter 6 Class Notes
Intermediate Algebra, MAT1033C
SI Leader Joe Brownlee
Palm Beach State College
Class Notes
6.1
Professor Burkett
SI Leader Joe Brownlee
6.1 – Greatest Common Factors and Factoring by Grouping
Writing a polynomial as the product of two or more simpler polynomials is called factoring.
Example 1:
12 + 24𝑧 may be written as 12(1 + 2𝑧).
The first step in factoring a polynomial is finding the greatest common factor, or GCF, of both the coefficients/constants
as well as the variables.
Example 2:
8𝑥 + 12
The GCF is 4, because 4 is the greatest factor that divides into both 8x and 12.
4(2𝑥 + 3)
Example 3:
9𝑧 − 18
To help us better see the GCF, let’s rewrite the above binomial: 9 ∙ 𝑧 − 9 ∙ 2
Since both terms have a 9, that will be our GCF; let’s factor it out to get: 9(𝑧 − 2)
We can check ourselves by using the distributive property: 9(𝑧 − 2)to get our original binomial of 9𝑧 − 18
Example 4:
Therefore, when we rewrite the original given trinomial, factoring out the GCF, we get 5𝑥(6𝑥 2 + 𝑥 − 5).
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Class Notes
6.1
Professor Burkett
SI Leader Joe Brownlee
Example 5:
Sometimes you will be given a polynomial where the leading coefficient is a negative. For these, you will try to find a
GCF, like always, but you will also factor out the negative so that the leading coefficient becomes positive.
−𝑎3 + 3𝑎2 − 5𝑎
Here we see the only factor each term has in common is an a, but we also see that the leading coefficient is negative, -.
So instead of simply factoring out an a, we’ll factor out –a so that when we rewrite our polynomial, we’ll have a positive
leading coefficient inside our parenthesis.
−𝑎(𝑎2 − 3𝑎 + 5)
Factoring by Grouping
When a polynomial has four terms, we “factor by grouping.” We do this by grouping the terms in pairs so that each pair
has a common factor.
STEP 1: GROUP TERMS. Collect the terms into groups so that each group has a common factor.
STEP 2: FACTOR WITHIN THE GROUPS. Factor out the common factor in each group.
STEP 3: FACTOR THE ENTIRE POLYNOMIAL. If each group now has a common factor, factor it out. If not, try a different
grouping.
Always check the factored form by multiplying (apply distributive property).
Example 6:
If we’re given the polynomial 2𝑤𝑥 + 10𝑤 + 7𝑥 + 35 and asked to factor, the first thing we’ll notice when we go to look
for a GCF is that there isn’t one, besides 1, between every term. We also see that this polynomial has four terms, and
whenever we see four terms, we know to factor by grouping.
So group the first two terms together, then factor out a GCF.
Then group the last two terms together, then factor out a GCF.
The final step is writing
our factors. So my first
factor is what’s inside
the parenthesis. The
second factor is what
was left outside the
parenthesis.
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When you factor by
grouping, what’s inside
the parenthesis has to
match; if it doesn’t, you’ve
done something wrong,
check your work.
Class Notes
6.2
Professor Burkett
SI Leader Joe Brownlee
6.2 – Factoring Trinomials
While there are other ways to factor trinomials, the easiest and quickest way is to use “the scissors” method.
Example 1:
Factor 𝑝2 + 6𝑝 + 5
Step 1: Factors. Look at the factors of the leading coefficient, in this case 1, and the constant, in this case 5,
and put them in a chart:
1
1,1
5
1,5
Step 2: Scissors. From this chart, we’re going to set up our scissors to determine the factors of the given
trinomial.
2
1
+
1
1*1=1
5
1*5=5
Multiply
1
+
When writing your final
factors, make sure you insert
the original variable in this
spot for each factor. In this
case, it was a p.
𝑝 + 6𝑝 + 5
Now that I’ve multiplied by scissors and came up with 1
and 5, I have to ask myself if there is a way to add the 1
and 5 together to get a postive 6? If so, the larger
number, in this case 5, gets the middle sign, +. And since
the last sign in this trinomial is a +, we know the other
factor will also have a + sign.
And there are your two factors of this trinomial:
(𝑝 + 1)(𝑝 + 5)
Throughout this chapter, when you’re factoring remember the questions you have to ask yourself on every
single problem:
1) IS THERE A GCF? If so, factor it out (this includes factoring out a -1 if necessary).
2) HOW MANY TERMS ARE THERE? (If there are two, look for the difference of squares or a problem with
cubes; if there are three terms, use scissors; if there are four terms, factor by grouping).
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Class Notes
6.2
Professor Burkett
SI Leader Joe Brownlee
Example 2:
Factor 𝑏2 − 7𝑏 + 10
Step 1: Factors. Look at the factors of the leading coefficient, in this case 1, and the constant, in this case 10,
and put them in a chart:
1
1,1
10
1,10
2,5
Step 2: Scissors. From this chart, we’re going to set up our scissors to determine the factors of the given
trinomial. Start with factors closest together, and if those don’t work, move out. In this case, 2 and 5 are
closer together than 1 and 10, so let’s start with 2 and 5.
1
-
2
1*2=2
Multiply
1
5
-
1*5=5
𝑏 2 − 7𝑏 + 10
Now that I’ve multiplied by scissors and came up with 2
and 5, I have to ask myself if there is a way to add the 2
and 5 together to get a negative 7? If so, the larger
number, in this case 5, gets the middle sign, -. And since
the last sign in this trinomial is a +, we know the other
factor will also have a - sign.
And there are your two factors of this trinomial:
(𝑏 − 2)(𝑏 − 5)
If for some reason your first attempt at scissors does not result in factors that work, you must try different
combinations until one is successful. For example, the below shows the various possibilities you can have
tried for the above problem, using the factors from the table we created in Step 1:
1
2 or 5 or 10 or 1
Multiply
1
5 or 2 or 1 or 10
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Class Notes
6.2
Professor Burkett
SI Leader Joe Brownlee
Example 3:
Factor −6𝑟 2 + 13𝑟 + 5
Step 1: Factor out a negative. If your leading coefficient is a negative, you must factor out a -1 (or a negative
GCF) before you begin to factor the trinomial.
-(6𝑟 2 − 13𝑟 − 5)
Now that my trinomial inside the parenthesis is positive, I can factor like normal.
Step 2: Factors. Look at the factors of the leading coefficient, in this case 6, and the constant, in this case 5,
and put them in a chart:
6
1,6
2,3
5
1,5
Step 3: Scissors. From this chart, we’re going to set up our scissors to determine the factors of the given
trinomial. Start with factors closest together, and if those don’t work, move out. In this case, 2 and 3 are
closer together than 1 and 6, so let’s start with 2 and 3.
2
-
5
3*5=15
Multiply
3
+
1
2*1=2
2
−(6𝑟 − 13𝑟 − 5)
Now that I’ve multiplied by scissors and came up with 15
and 2, I have to ask myself if there is a way to subtract
the 15 and 2 to get a negative 13? If so, the larger
number, in this case 15, gets the middle sign, -. And
since the last sign in this trinomial is a -, we know the
other factor will have the opposite sign, a +.
And there are your two factors of this trinomial:
−(2𝑟 − 5)(3𝑟 + 1)
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Class Notes
6.3
6.3 Special Factoring
Special factoring will fall under one of the following circumstances:
Let’s practice some of the ones from above:
Example 1: Difference of Squares
Factor the polynomial.
𝑝2 − 100
= 𝑝2 − 102
100 = 102
= (𝑝 + 10)(𝑝 − 10)
Factor the difference of squares.
Example 2: Difference of Squares
Factor the polynomial.
4𝑎2 − 64
= 4(𝑎2 − 16)
Factor out the GCF, 4.
= 4(𝑎 + 4)(𝑎 − 4)
Factor the difference of squares.
Example 3: Difference of Squares
Factor the polynomial.
𝑥 4 − 81
= (𝑥 2 + 9)(𝑥 2 − 9)
Factor the differences of squares.
= (𝑥 2 + 9)(𝑥 + 3)(𝑥 − 3)
Notice(𝑥 2 − 9). Factor the differences of squares again.
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Professor Burkett
SI Leader Joe Brownlee
Class Notes
6.3
Professor Burkett
SI Leader Joe Brownlee
Example 4: Difference of Cubes
Factor the polynomial.
𝑥 3 − 1000
= (𝑥)3 − (10)3
Difference of cubes.
= (𝑥 − 10)(𝑥 2 + 10𝑥 + 100)
Factor.
Example 5: Difference of Cubes
Factor the polynomial.
8𝑘 3 − 𝑦 3
= (2𝑘)3 − (𝑦)3
Difference of cubes.
Multiply the two binomial terms together.
= (2𝑘 − 𝑦)(4𝑘 2 + 2𝑘𝑥 + 𝑦 2 )
Square it.
Factor.
Square it.
Example 6: Sum of Cubes
Factor the polynomial.
27𝑚3 + 125𝑛3
= (3𝑚)3 + (5𝑛)3
Sum of cubes.
= (3𝑚 + 5𝑛)(3𝑚2 − 15𝑚𝑛 + 25𝑛2 )
Factor.
Example 7: Sum of Cubes.
Factor the polynomial.
(𝑎 − 4)3 + 𝑏3
= (𝑥)3 + (𝑏)3
Sum of cubes. Choose a single variable to take the place of (a-4); we’ll choose x.
= (𝑥 + 𝑏)(𝑥 2 − 𝑥𝑏 + 𝑏2 )
Factor.
= (𝑎 − 4 + 𝑏)[(𝑎 − 4)2 − (𝑎 − 4)𝑏 + 𝑏2 ]
= (𝑎 − 4 + 𝑏)(𝑎2 − 8𝑎 + 16 − 𝑎𝑏 + 4𝑏 + 𝑏2 )
Plug your original expression back in for x.
Multiply and combine like terms where possible.
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Class Notes
6.4
Professor Burkett
SI Leader Joe Brownlee
6.4 – A General Approach to Factoring
Section 6.4 introduced no additional content, it was simply a combination review of 6.1, 6.2, and 6.3.
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Class Notes
6.5
Professor Burkett
SI Leader Joe Brownlee
6.5 – Solving Equations by Factoring
In the previous sections, we worked with polynomials in order to get their factors. In this chapter, we’ll be
doing the same process, factoring, but when we get the factors of our polynomials, however many, we’ll set
each one to 0 and solve for the variable. Let’s see how that works.
Example 1:
Solve each equation.
(3𝑥 + 5)(𝑥 + 1) = 0
Use the zero-factor property on the given factors by setting each to 0 and solving.
3𝑥 + 5 = 0
𝑥+1=0
3𝑥 = −5
𝑥 = −1
𝑥=−
5
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Example 2:
Solve each equation.
(3𝑥 + 11)(5𝑥 − 2) = 0
Use the zero-factor property on the given factors by setting each to 0 and solving.
3𝑥 + 11 = 0
5𝑥 − 2 = 0
3𝑥 = −11
5𝑥 = 2
𝑥=−
11
3
𝑥=
2
5
Example 3:
Solve each equation.
3𝑥 2 − 𝑥 = 4
3𝑥 2 − 𝑥 − 4 = 0 Subtract 4 from both sides and set the equation to 0 by writing it in standard form.
(𝑥 + 1)(3𝑥 − 4) = 0
𝑥+1=0
𝑥 = −1
Factor the trinomial. See 6.2.
3𝑥 − 4 = 0
𝑥=
Set each factor to 0 and solve for x.
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3
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Class Notes
6.5
Professor Burkett
SI Leader Joe Brownlee
Example 4:
Solve.
12𝑥 = 2𝑥 3 + 5𝑥 2
2𝑥 3 + 5𝑥 2 − 12𝑥 = 0 Subtract 12x from both sides and set the equation to 0 by writing it in standard form.
𝑥(2𝑥 2 + 5𝑥 − 12) = 0
Factor out the GCF, in this case an x.
𝑥(𝑥 + 4)(2𝑥 − 3) = 0
Factor the trinomial.
𝑥=0
𝑥+4 =0
2𝑥 − 3 = 0
𝑥=0
𝑥 = −4
𝑥=
Set each factor, including your GCF, to zero and solve.
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Note: If your GCF is only a number, when you set it equal to 0 you get a false statement; therefore you drop
that as an answer. For example, 3(𝑥 + 4)(6𝑥 − 2) = 0. When you set 3 equal to 0 you get 3≠0, so drop that
as an answer and solve for the remaining factors.
Example 5:
Solve.
(𝑥 + 6)(𝑥 − 2) = −8 + 𝑥
𝑥 2 + 4𝑥 − 12 = −8 + 𝑥
𝑥 2 + 3𝑥 − 4 = 0
(𝑥 + 4)(𝑥 − 1) = 0
FOIL the left side of the equation.
Set the equation equal to 0 by adding 8 to both sides, then subtracting x from both sides.
Factor the trinomial.
𝑥+4=0
𝑥−1=0
𝑥 = −4
𝑥=1
Set each factor equal to zero and solve.
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