9.8
TRANSCRIPT
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9.8 Factor Special Products
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Difference of Two Squares (Factor)
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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 –
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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.
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Perfect Square Trinomial
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Perfect Square Trinomial (Factor)
Algebra
Algebra
Example
Example
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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++
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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+
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ANSWER The correct answer is D.
Multiple Choice PracticeExample 3
x 6y( )23– –= Perfect square trinomial pattern
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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
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Solve a polynomial equationExample 4
–3
1ANSWER The solution of the equation is .
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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.
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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.
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Solve a vertical motion problemExample 5
ANSWER
The sponge lands on the ground 2 seconds after it is dropped.
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9.8 Warm-Up (Day 1)Factor the polynomial.
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3.
4.
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9.8 Warm-Up (Day 2)Solve the equation.
1.
2.
3.