Factor.

Warm-Up Exercises

1. x 2 2x+ 48– ANSWER ( )8+x ( )6x –

2. 9y 2 27y 20+– ANSWER ( )43y –( )53y –

3. 2y 2 16y 32+– ANSWER ( )4y –( )4–y2

4. The sides of a square are (2s 3) inches. Find the area.

+

ANSWER 4s 2 12s 9+–( in.2)

Example 1 Factor a Difference of Two Squares

Factor the expression.

a. m 2 – 25 b. q 2 – 625 c. 9y 2 – 16

Difference of two squares pattern

= ( )5+m ( )5m –

SOLUTION

Write as .a. m 2 – 25 = m 2 – 52 a 2 – b 2

Difference of two squares pattern

= ( )25+q ( )25q –

Write as .b. = q 2 – 252 a 2 – b 2q 2 – 625

Example 1 Factor a Difference of Two Squares

Write as .a 2 – b 2c. 9y 2 – 16 = – 42( )23y

Difference of two squares pattern

= ( )4+3y ( )43y –

Checkpoint

1.

Factor the expression.

Factor Using the Difference of Two Squares Pattern

x 2 – 36 ANSWER ( )6+x ( )6x –

2. r 2 – 100 ANSWER ( )10+r ( )10r –

3. 9m 2 – 64 ANSWER ( )8+3m ( )83m –

4. p 2 – 8141 ANSWER p +

21 9 p

21 9–

Example 2 Factor a Perfect Square Trinomial

Factor the expression.

m 2 16m + 64+

16r 2 56r + 49–

a. b. 9p 2 30p + 25+

c.

( )28m += Perfect square trinomial pattern

SOLUTION

m 2 16m + 64+ = m 2 2+ ( )m ( )8 + 82a. Write asa 2 2ab ++ b 2.

Example 2 Factor a Perfect Square Trinomial

Perfect square trinomial pattern

( )253p +=

( )274r= – Perfect square trinomial pattern

c. 16r 2 56r + 49– = ( )7 + 72( )24r 2( )4r– Write asa 2 2ab + b 2– .

b. 9p 2 30p + 25+ = ( )5 + 52( )23p 2( )3p+ Write asa 2 2ab ++ b 2.

Example 3 Factor Out a Common Constant

Factor .5u 2 40u + 80–

= ( )24u –5 Perfect square trinomial pattern

SOLUTION

5u 2 40u + 80– = ( )u 2 8u + 16–5 Factor out 5.

( )4 + 42u 2 2– ( )u[ ]= 5 Write as a 2 2ab + b 2– .

CHECK Check your answer by multiplying.

( )4u –=( )24u –5 5 ( )4u – = ( )u 2 8u + 16–5

= 5u 2 40u + 80–

Checkpoint

Factor the expression.

Factor the Expression

5. x 2 10x + 25+ ANSWER ( )25x +

4x 2 12x + 9+6.

7. 9p 2 24p + 16–

ANSWER ( )232x +

ANSWER ( )243p –

ANSWER ( )21x +58. 5x 2 10x + 5+

Checkpoint

Factor the expression.

Factor the Expression

9. 8y 2 18–

10. 12u 2 36u + 27–

ANSWER ( )32y –( )3+2y2

ANSWER ( )232u –3

Example 4 Solve a Quadratic Equation

Solve –9p 2 = 25.+ 30p

SOLUTION

Write original equation.–9p 2 = 25+ 30p

Write in standard form.9p 2 = 0+ 30p + 25

( )2 Write as3p = 0+ 2 + 52( )3p ( )5 a 2 + 2ab + b 2.

Perfect square trinomial pattern= 0( )23p + 5

Use the zero product property.= 03p + 5

Solve for p.=p35–

Example 4 Solve a Quadratic Equation

ANSWER The solution is .35–

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 180C 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.

=

SOLUTIONSTEP 1 Find the circumference of a rope that can be

used to lift 720 pounds.

= 180C 2S Function for safe working load

Example 5 Use a Quadratic Equation as a Model

= 180C 2720 Substitute 720 for S.

= 180C 20 Write in standard form.– 720

= 1800 Factor out 180.– 4( )C 2

= 1800 Difference of two squares pattern+ 2( )C – 2( )C

= 0+ 2C or = 02C – Use the zero product property.

= 2 or = 2C Solve for C.C –

Reject the negative value of C. The rope must have a circumference of 2 inches.

Example 5 Use a Quadratic Equation as a Model

Substitute 2 for C in C= πd2 = πd.

Solve for d.=d2π ≈ 0.637

ANSWERThe rope must have a diameter of or about 0.637 inch.2

π

STEP 2 Find the diameter of the rope. Use the formula for circumference, C = πd.

Checkpoint

ANSWER 3

ANSWER 10

11.

96xx 2 =

Solve the equation.

Solve a Quadratic Equation

– –

12.

124a 2 =– ANSWER 4, 4–

13. 20040y2y 2 = –