1. Warm-Up Exercises
Factor. 1.
x 2 2x + 48 –
ANSWER ( ) 8 + x 6 – 2.
9y 2
27y 20 + –
ANSWER ( ) 4 3y – 5 3.
2y 2
16y 32 + –
ANSWER ( ) 4 y – 2
The sides of a square are (2s 3) inches. Find the
area. +
ANSWER
4s 2
12s 9 + – (
in.2 )

2. Example 1 Factor the expression. a. m 2 – 25 b. q 2 – 625 c. 9y 2 – 16
Factor a Difference of Two Squares
Factor the expression. a.
m 2 – 25 b.
q 2 –
625 c.
9y 2 – 16
SOLUTION
Write as a.
m 2 – 25 = 52
a 2
b 2
Difference of two squares
pattern = ( ) 5 + m –
Write as b. =
q 2 –
252
a 2
b 2
625
Difference of two squares
pattern = ( ) 25 + q – 2

3. Example 1 c. 9y 2 – 16 = 42 ( )2 3y = ( ) 4 + 3y –
Factor a Difference of Two Squares c.
9y 2 – 16 = 42 ( )2 3y
Write as
a 2 –
b 2
Difference of two squares
pattern = ( ) 4 + 3y –

4. Checkpoint Factor the expression. 1. x 2 – 36 ANSWER ( ) 6 + x – 2.
Factor Using the Difference of Two Squares
Pattern
Factor the expression. 1.
x 2 – 36
ANSWER ( ) 6 + x – 2.
r 2 –
100
ANSWER ( ) 10 + r – 3.
9m 2 – 64
ANSWER ( ) 8 + 3m – 4.
p 2 – 81 4 1
ANSWER p + 2 1 9 –

5. Example 2 Factor the expression. a. m 2 16m + 64 b. 9p 2 30p + 25 c.
Factor a Perfect Square Trinomial
Factor the expression. a.
m 2
16m + 64 b.
9p 2
30p + 25 c.
16r 2
56r + 49 –
SOLUTION
m 2
16m + 64 = 2 ( ) m 8 82 a.
Write as
a 2
2ab
b 2 . ( )2 8 m + =
Perfect square trinomial pattern

6. Example 2 b. 9p 2 30p + 25 = ( ) 5 52 )2 3p 2 ( )2 5 3p + = c. 16r 2
Factor a Perfect Square Trinomial b.
9p 2
30p + 25 = ( ) 5 52 )2 3p 2
Write as
a 2
2ab
b 2 .
Perfect square trinomial pattern ( )2 5 3p + = c.
16r 2
56r + 49 – = ( ) 7 72 )2 4r 2
Write as
a 2
2ab
b 2 . ( )2 7 4r = –
Perfect square trinomial pattern 6

7. [ ] Example 3 Factor . 5u 2 40u + 80 – SOLUTION 5u 2 40u + 80 – = ( )
Factor Out a Common Constant
Factor
5u 2
40u + 80 –
SOLUTION
5u 2
40u + 80 – = ( )
u 2 8u 16 5
Factor out 5. ( ) 4 + 42
u 2 2 – u [ ] = 5
Write as
a 2
2ab
b 2 . = ( )2 4 u – 5
Perfect square trinomial pattern
CHECK
Check your answer by multiplying. ( ) 4 u – = )2 5
u 2 8u + 16
5u 2
40u 80 7

8. Checkpoint Factor the expression. 5. x 2 10x + 25 ANSWER ( )2 5 x +9 6.
ANSWER ( )2 3 2x + 7.
9p 2
24p + 16 –
ANSWER ( )2 4 3p – 8.
5x 2
10x + 5
ANSWER ( )2 1 x + 5

9. Checkpoint Factor the expression. 9. 8y 2 18 – ANSWER ( ) 3 2y – + 2
10.
12u 2
36u + 27 –
ANSWER ( )2 3 2u –

10. Example 4 Solve – 9p 2 = 25. + 30p SOLUTION – 9p 2 = 25 + 30p 9p 2 = +
Solve a Quadratic Equation
Solve –
9p 2 =
25. +
30p
SOLUTION
Write original equation. –
9p 2 = 25 +
30p
Write in standard form.
9p 2 = +
30p 25 ( )2
Write as 3p = + 2 52 ) 5
a 2
2ab
b 2.
Perfect square trinomial pattern = ( )2 3p + 5
Use the zero product property. = 3p + 5
Solve for p. = p 3 5 – 10

11. Example 4
Solve a Quadratic Equation
ANSWER
The solution is 3 5 – 11

12. Find the circumference of a rope that can be used to lift 720 pounds.
Example 5
Use a Quadratic Equation as a Model
Rope Strength Every rope has a safe working load. A rope should not be used to lift a weight greater than its safe working load. The safe working load S (in pounds) for a rope can be found using the function S C 2 where C is the circumference (in inches) of the rope. Find the diameter of rope needed to lift an object that weighs 720 pounds. =
SOLUTION
STEP 1
Find the circumference of a rope that can be used to lift 720 pounds. =
180C 2 S
Function for safe working load 12

13. Example 5 720 = 180C 2 = 180C 2 – 720 = 180 – 4 ( ) C 2 = 180 + 2 ( )
Use a Quadratic Equation as a Model
720 =
180C 2
Substitute 720 for S. =
180C 2
Write in standard form. –
720 =
180
Factor out 180. – 4 ( )
C 2 =
180
Difference of two squares pattern + 2 ( ) C – = + 2 C or –
Use the zero product property. = 2 or C
Solve for C. –
Reject the negative value of C. The rope must have a circumference of 2 inches. 13

14. Find the diameter of the rope. Use the formula for circumference, C =
Example 5
Use a Quadratic Equation as a Model
STEP 2
Find the diameter of the rope. Use the formula for circumference, C =
πd.
Substitute 2 for C in C = πd 2
πd.
Solve for d. = d 2 π ≈
0.637
ANSWER
The rope must have a diameter of or about inch. 2 π

15. Checkpoint Solve the equation. 11. x 2 – ANSWER 3 6x = – 9 12. 12 4
Solve a Quadratic Equation
Solve the equation.
11.
x 2 –
ANSWER 3 6x = – 9
12. 12 4
a 2 = –
ANSWER
4, 4 –
13.
200
40y
2y 2 = –
ANSWER 10