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```5-4 Factoring
M11.A.1.2.1: Find the Greatest Common Factor and/or the Least
Common Multiple for sets of monomials
M11.D.2.1.5: Solve quadratic equations using factoring
M11.D.2.2.2: Factor algebraic expressions, including differences of
squares and trinomials
Objectives
Finding Common Binomial Factors
Factoring Special Expressions
Vocabulary
 Factoring is rewriting an expression as the product
of its factors.
 The greatest common factor (GCF) of an
expression is a common factor of the terms of the
expression.
Finding Common Factors
Factor each expression.
a. 15x2 + 25x + 100
15x2 + 25x + 100 = 5(3x2) + 5(5x) + 5(20)
= 5(3x2 + 5x + 20)
Factor out the GCF, 5
Rewrite using the
Distributive Property.
b. 8m2 + 4m
8m2 + 4m = 4m(2m) + 4m(1)
= 4m(2m + 1)
Factor out the GCF, 4m
Rewrite using the
Distributive Property.
Factoring when ac > 0 and b > 0
Factor x2 + 10x + 24.
Step 1: Find factors with product ac and sum b.
Since ac = 24 and b = 10, find positive factors with product 24
and sum 11.
Factors of 24
Sum of factors
1, 24
25
2, 12
14
3, 8
11
6, 4
10
}
}
Step 2: Rewrite the term bx using the factors you found. Group the
remaining terms and find the common factors for each group.
x2 + 10x + 24
x2 + 4x + 6x + 24
Rewrite bx : 10x = 4x + 6x.
x(x + 4) + 6(x + 4)
Find common factors.
Continued
(continued)
Step 3: Rewrite the expression as a product of two binominals.
x(x + 4) + 6(x + 4)
(x + 6)(x + 4)
Rewrite using the Distributive Property.
Check: (x + 6)(x + 4) = x2 + 4x + 6x + 24
= x2 + 10x + 24
Factoring when ac > 0 and b < 0
Factor x2 – 14x + 33.
Step 1: Find factors with product ac and sum b.
Since ac = 33 and b = –14, find negative factors with product 33
and sum b.
Factors of 33
Sum of factors
–1, –33
–34
–3, –11
–14
Step 2: Rewrite the term bx using the factors you found. Then find common
factors and rewrite the expression as a product of two binomials.
x2 + 14x + 33
x2 – 3x – 11x + 33Rewrite bx.
}
}
x(x – 3) – 11(x – 3)
Find common factors.
(x – 11)(x – 3)
Rewrite using the Distributive
Property.
Factoring When ac < 0
Factor x2 + 3x –28.
Step 1: Find factors with product ac and sum b.
Since ac = –28 and b = 3, find factors 2 with product –28 and
sum 3.
Factors of –28
Sum of factors
1, –28
–27
–1, 28
27
2, –14
–12
–2, 14
12
4, –7
–3
–4, 7
3
Step 2: Since a = 1, you can write binomials using the factors you found.
x2 + 3x – 28
(x – 4)(x + 7)
Use the factors you found.
Factoring When a ≠ 1 and ac > 0
Factor 6x2 – 31x + 35.
Step 1: Find factors with product ac and sum b.
Since ac = 210 and b = –31, find negative factors with product
210 and sum –31.
Factors of 210
Sum of factors
–1, –210 –2, –105 –3, –70 –5, –42 –10, –21
–211
–107
–73
–47
–31
Step 2: Rewrite the term bx using the factors you found. Then find common
factors and rewrite the expression as the product of two binomials.
6x2 – 31x + 35
2x(3x – 5) – 7(3x – 5)
Find common factors.
(2x – 7)(3x – 5)
Rewrite using the Distributive Property.
}
Rewrite bx.
}
6x2 – 10x – 21x + 35
Factoring When a ≠ 1 and ac < 0
Factor 6x2 + 11x – 35.
Step 1: Find factors with product ac and sum b.
Since ac = 210 and b = 11, find factors with product –210 and
sum 11.
Factors of –210 –1, –210
Sum of factors
–209
Factors of –210
Sum of factors
–3, 70
67
–1, 210
209
5, –42
–37
2, –105 –2, 105
–103
103
–5, 42
37
10, –21
–11
3, –70
–67
–10, 21
11
Step 2: Rewrite the term bx using the factors you found. Then find common
factors and rewrite the expression as the product of two binomials.
6x2 + 11x + 35
6x2 – 10x + 21x – 35
Rewrite bx.
2x(3x – 5) + 7(3x – 5)
Find common factors.
(2x + 7)(3x – 5)
Rewrite using the Distributive Property.
Vocabulary
A perfect square trinomial is the product you obtain when you
square a binomial.
a² + 2ab + b² = (a + b)²
a² - 2ab + b² = (a – b)²
Factoring a Perfect Square
Trinomial
Factor 100x2 + 180x + 81.
100x2 + 180x + 81 = (10x)2 + 180 + (9)2
Rewrite the first and third
terms as squares.
= (10x)2 + 180 + (9)2
Rewrite the middle term to
verify the perfect square
trinomial pattern.
= (10x + 9)2
a2 + 2ab + b2 = (a + b)2
Real World Example
A square photo is enclosed in a square frame, as shown in
the diagram. Express the area of the frame (the shaded area) in
completely factored form.
Relate: frame area equals the outer area minus the inner area
Define: Let x = length of side of frame.
Write:
area = x2 – (7)2
= (x + 7)(x – 7)
The area of the frame in factored form is (x + 7)(x – 7) in2.
Homework
Pg 263 # 1, 2, 7, 13, 19, 25, 31
Pg 264 # 37, 38, 46
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