Page 138 - Math Course 2 (Book 1)

Factoring Binomials

Mo. 4

2
2
Lesson 5 Factor 16y – 81z .
2
2
2
2
2
16y – 81z = (4y) – (9z) 16y = 4y • 4y and
2
81z = 9z • 9z
KEY CONCEPTS:
1. Factor binomials that are differences of = (4y + 9z)(4y – 9z) Factor the
squares. difference of
2. Solve equations of the form x2 + bx + c. squares.
Answer (4y + 9z)(4y – 9z)

MO. 4 - L5a

Binomial Difference of Factor 3b – 27b.
3
Squares If the terms of a binomial have a common factor,
the GCF should be factored out f rst before trying to
apply any other factoring technique.
Key Concept 3b – 27b = 3b(b – 9) The GCF of 3b
2
2
3
and 27b is 3b.
Difference of Squares = 3b[ b – 3 ] b = b • b and
2
2
2
9 = 3 • 3
Symbols a – b = (a + b)(a – b) or (a – b)(a + b)
2
2
= 3b(b + 3)(b – 3) Factor the
difference of
2
Examples x – 9 = (x + 3)(x – 3) or (x – 3)(x + 3)
squares.
Answer 3b(b + 3)(b – 3)

Let’s Begin Apply a Factoring Technique More
Than Once

Example

Factor the Difference of Squares Factor y – 625.
4

Examples y – 625. Original binomial
4
2
4
2
2 2
2
= [(y ) – 25 ] y = y • y and
2
Factor m – 64. 625 = 25 • 25
2
2
2
2
m – 64 = m – 8 2 Write in the form = (y + 25)(y – 25) Factor the difference of
2
2
a – b . squares.
2
2
2
2
= (m + 8)(m – 8) Factor the = (y + 25)(y – 5 ) y = y • y and 25 = 5 • 5
difference of
2
squares. = (y + 25)(y + 5)(y – 5) Factor the difference of
squares.
2
Answer (m + 8)(m – 8) Answer (y + 25)(y + 5)(y – 5)
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