9.8 Factor Special Products

Difference of Two Squares (Factor)

Factor the difference of two squares

Factor the polynomial.

Example 1

3625m2a. –

( )6+5m ( )65m – Difference of two squares pattern

=

Write as .625m –= ( )2 b2a2 –

49y2x2b. – Write as .x2 –= b2a2 –7y( )2

( )7y+x Difference of two squares pattern

= ( )7yx –

18n28c. – Factor out common factor.= ( )9n242 –

Write as .= 2[ ]22 – 3n( )2 9n24 – b2a2 –

Factor the difference of two squaresExample 1

( )3n+2 Difference of two squares pattern

= ( )3n2 –2

+9– 4x2d.

Difference of two squares pattern

( )3+2x ( )32x –=

Write as .322x –= ( )2 b2a2 –

94x2 –= Rewrite as difference.

Perfect Square Trinomial

Perfect Square Trinomial (Factor)

Algebra

Algebra

Example

Example

Factor perfect square trinomials

Factor the polynomial.

Example 2

12nn2a. – 36+ Write as .= ( )2n2 – 62+6n • 2aba2 – b2+

( )26n – Perfect square trinomial pattern

=

12x9x2b. – 4+

( )223x – Perfect square trinomial pattern

=

Write as .= ( )2– 22+23x • 2aba2 – b2+3x( )2

( )2t2s + Perfect square trinomial pattern

=

Write as .= ( )2+ t2+t2s • 2aba2 + b2+2s( )24st4s2c. t2++

What is the factored form of ?

Multiple Choice PracticeExample 3

36xy3x2 108y2+– –

3x 6y–( )2 x 6y( )23– –

3x 6y– –( )2 x 6y+( )23–

SOLUTION

36xy3x2 108y2+– – 12xyx2 – 36y2+3– ( )=

Factor out .3–

( )2– +6yx •x2 6y( )2[ ]3–=

Write as .12xyx2 – 36y2+ 2aba2 – b2+

ANSWER The correct answer is D.

Multiple Choice PracticeExample 3

x 6y( )23– –= Perfect square trinomial pattern

Solve a polynomial equationExample 4

Solve the equation .x 0x2 =+3

2+9

1

Write original equation.x 0x2 =+3

2+9

1

Multiply each side by 9.6x 09x2 =+ + 1

( )213x + Perfect square trinomial pattern0=

13x + Zero-product property0=

Solve for x.x = –3

1

Write left side as .2aba2 + b2+( )2+ 1+13x •3x( )2 0=( )2

Solve a polynomial equationExample 4

–3

1ANSWER The solution of the equation is .

Example 5

A window washer drops a wet sponge from a height of 64 feet. After how many seconds does the sponge land on the ground?

FALLING OBJECT

Solve a vertical motion problem

SOLUTION

Use the vertical motion model to write an equation for the height h (in feet) of the sponge as a function of the time t (in seconds) after it is dropped.Because the sponge was dropped, it has no initialvertical velocity. To determine when the sponge lands on the ground, find the value of t for which the height is 0.

Solve a vertical motion problemExample 5

Vertical motion modelvt16t2 + + s=h –

0( ) Substitute 0 for h, 0 for v, and 64 for s.

t16t2 + + 64=0 –

Factor out .=0 16– ( )4t2 – 16–

Difference of two squares pattern=0 16– ( )2+t ( )2–t

Zero-product property0=2+t 2–t 0=or

= Solve for t.t 2=2– tor

Disregard the negative solution of the equation.

Solve a vertical motion problemExample 5

ANSWER

The sponge lands on the ground 2 seconds after it is dropped.

9.8 Warm-Up (Day 1)Factor the polynomial.

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2.

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4.

9.8 Warm-Up (Day 2)Solve the equation.

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2.

3.