1.
Difference of Two Squares (Factor)
a - b = (a + b)(a - b)
2
2
y -16 = y - 4 = (y + 4)(y - 4)
2
2
2
2.
Example 1
Factor the difference of two squares
Factor the polynomial.
a. 25m2 – 36 = ( 5m )2 – 62
Write as a2 – b2.
= (5m + 6 ) (5m – 6 ) Difference of two squares
pattern
b. x2 – 49y2 = x2 – ( 7y)2
= (x + 7y ) (x – 7y)
c. 8 – 18n2 = 2( 4 – 9n2 )
= 2[22 – ( 3n)2]
Write as a2 – b2.
Difference of two squares
pattern
Factor out common factor.
Write 4 – 9n2 as a2 – b2.
3.
Example 1
Factor the difference of two squares
= 2(2 + 3n ) (2 – 3n)
d. – 9 + 4x2 = 4x2 – 9
Difference of two squares
pattern
Rewrite as difference.
= ( 2x )2 – 32
Write as a2 – b2.
= (2x + 3 ) (2x – 3 )
Difference of two squares
pattern
4.
Perfect Square Trinomial
x 2 + 6x + 9
a + 2ab + b
2
x 2 -10x + 25
2
a - 2ab + b
2
2
5.
Perfect Square Trinomial (Factor)
a2 + 2ab + b2 = (a + b)2
Algebra
Example
x + 6x + 9 = x + 2(x ×3) + 3 = (x + 3)
Algebra
(a - b)2
a - 2ab + b =
2
2
2
2
2
2
(x - 5)2
Example x -10x + 25 = x - 2(x × 5)+ 5 =
2
2
2
6.
Example 2
Factor perfect square trinomials
Factor the polynomial.
a. n2 – 12n + 36 = n2 – 2( n • 6) + 62
= (n – 6 )2
Write as a2 – 2ab + b2.
Perfect square
trinomial pattern
b. 9x2 – 12x + 4 = ( 3x)2 – 2( 3x • 2) + 22 Write as a2 – 2ab + b2.
= (3x – 2 )2
Perfect square
trinomial pattern
c. 4s2 + 4st + t2 = ( 2s )2 + 2( 2s • t ) + t2 Write as a2 + 2ab + b2.
= (2s + t )2
Perfect square
trinomial pattern
7.
Example 3
Multiple Choice Practice
What is the factored form of – 3x2 + 36xy – 108y2 ?
( – 3x – 6y )2
– 3( x + 6y )2
( 3x – 6y )2
– 3( x – 6y )2
SOLUTION
–3x2 + 36xy – 108y2 = –3( x2 – 12xy + 36y2 )
Factor out – 3 .
2
= –3[x2 – 2 (x • 6y ) + ( 6y) ]
Write x2 – 12xy + 36y2 as a2 – 2ab + b2.
8.
Example 3
Multiple Choice Practice
= – 3( x – 6y )2
ANSWER
Perfect square trinomial pattern
The correct answer is D.
9.
Example 4
Solve a polynomial equation
Solve the equation
x2
x2
1
x +
+
= 0.
3
9
= 0
Write original equation.
6x + 1 = 0
Multiply each side by 9.
2
1
x +
+
3
9x2 +
2
9
( 3x )2 + 2 ( 3x • 1 ) + ( 1 )2 = 0
Write left side as a2 + 2ab + b2.
( 3x + 1 )2 = 0
Perfect square trinomial pattern
3x + 1 = 0
x = –
Zero-product property
1
3
Solve for x.
10.
Example 4
ANSWER
Solve a polynomial equation
The solution of the equation is
–
1
3
.
11.
Example 5
Solve a vertical motion problem
FALLING OBJECT
A window washer drops a wet sponge from a height of
64 feet. After how many seconds does the sponge land
on the ground?
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 initial
vertical velocity. To determine when the sponge lands
on the ground, find the value of t for which the height is
0.
12.
Example 5
Solve a vertical motion problem
h = – 16t2 + vt + s
Vertical motion model
0 = – 16t2 + ( 0 ) t + 64
Substitute 0 for h, 0 for v, and
64 for s.
0 = – 16 ( t2 – 4 )
Factor out – 16 .
0 = – 16 ( t + 2 ) ( t – 2 )
Difference of two squares pattern
t + 2 = 0
or
t = –2
or
t – 2 = 0
t = 2
Zero-product property
Solve for t.
Disregard the negative solution of the equation.
13.
Example 5
Solve a vertical motion problem
ANSWER
The sponge lands on the ground 2 seconds after it is
dropped.
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