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Over Chapter 7 A. A B. B C. C D. D 5a 3 b
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Over Chapter 7
A. AB. BC. CD. D
–18x8y
Simplify (–2x4y)(–3x2)2.
Over Chapter 7
A. AB. BC. CD. D
9x2 – 12x + 4
Simplify (3x – 2)2.
Over Chapter 7
A. AB. BC. CD. D
4x2 – 36
Simplify (2x – 6)(2x + 6).
Over Chapter 7
A. AB. BC. CD. D
A. 12
B. 8
C. 6
D. 4
Which value of n makes (c6 – 3)2 = cn – 6c6 + 9 true?
• Factor monomials.
• Find the greatest common factors of monomials
Monomial in Factored Form
Factor 18x2y3 completely.
18x2y3 = 2 ● 9 ● x ● x ● y ● y ● y 18 = 2 ● 9, x2 = x ● x, and y3 = y ● y ● y
= 2 ● 3 ● 3 ● x ● x ● y ● y ● y 9 = 3 ● 3
Answer: 18x2y3 in factored form is 2 ● 3 ● 3 ● x ● x ● y ● y ● y.
A. AB. BC. CD. D
3 ● 5 ● a ● a ● a ● b ● b
Factor 15a3b2 completely.
GCF of a Set of Monomials
Find the GCF of 27a2b and 15ab2c.
27a2b = 3 ● 3 ● 3 ● a ● a ● bFactor each number.
Answer: The GCF of 27a2b and 15ab2c is 3 ● a ● b or 3ab.
15ab2c= 3 ● 5 ● a ● b ● b ● cCircle the common prime factors.
A. AB. BC. CD. D
13xy3
Find the GCF of 39x2y3 and 26xy4.
Find a GCF
GEOMETRY The lengths of the sides of a triangle are 12wz2, 8wz, and 16w2z. Find the GCF of the three lengths.
12wz2 = 3 ● 22 ● w ● z2
8wz = 23 ● w ● z 16w2z = 24 ● w2 ● z
The common prime factors are 22 ● w ● z.
Answer: So, the GCF is 4wz.
A. AB. BC. CD. D
4 bracelets
Mary is making bracelets with large and small beads. She has 20 large beads and 96 small beads. What is the greatest number of bracelets she can make with having any beads left over?
