5.6

Report
Section 5.6 Review Difference of Two Squares
Sum & Difference of Two Cubes
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Recognizing Perfect Squares
Difference of Two Squares
Recognizing Perfect Cubes
Sum of Two Cubes
Difference of Two Cubes
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Recognizing Perfect Squares (X)2
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Why? Because it enable efficient factoring!
Memorize the first 16 perfect squares of integers
1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256
12 22 32 42 52 62 72 82 92 102 112 122 132 142 152 162
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The opposites of those integers have the same square!
1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256
(-1)2 .. (-4)2 .. (-7)2 ..
(-10)2
..
(-15)2
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Variables with even exponents are also perfect squares
x2 = (x)2 y6 = (y3)2 y6 = (-y3)2 a2 b14 = (ab7)2
Monomials, too, if all factors are also perfect squares
a2 b14 = (ab7)2 81x8 = (9x4)2 225x4y2z22 = (15x2yz11)2
a2 b14 = (-ab7)2 81x8 = (-9x4)2 225x4y2z22 = (-15x2yz11)2
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… don’t forget those opposites!
2
The Difference between 2 Squares
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F2 – L2 factors easily to (F + L)(F – L)
Examine 49x2 – 16
(7x)2 – (4)2
(7x + 4)(7x – 4)
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Remember to remove common factors and to factor completely
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4U:
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64a4 – 25b2
(8a2)2 – (5b)2
(8a2 + 5b)(8a2 – 5b)
x4 – 1
(x2) – 12
2x4y – 32y
2y(x4 – 16)
(x2 + 1)(x2 – 1)
2y(x2 + 4)(x2 – 4)
2+4)(x+2)(x-2)
(x2+1)(x+1)(x-1)
2y(x
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Perfectly Square Practice
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p2 + q2 = prime! (sum of 2 “simple” squares is never factorable)
256x2 – 100 = (16x)2 – (10)2 = (16x + 10)(16x – 10)
256x2 – 100 = 4(64x2 – 25) = 4(8x + 5)(8x – 5)
16a2 – 11 = prime (middle term can’t disappear unless both are 2 )
x2 – (y + z)2 = (x + y + z)(x – y – z) (note that –(y+z)=–y–z)
x2 + 6x + 9 – z2 = (x + 3)2 – z2 = (x + 3 + z)( x + 3 – z)
3a4 – 3 = 3(a4–1) = 3(a2+1)(a2–1) = 3(a2+1)(a+1)(a–1)
Ready for Perfect Cubes?
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A Disappearing Act
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p2 – pq + q2
x
p+q
p2q – pq2 + q3
p3 – p2q + pq2
so, the sum is
p3
+ q3 = p 3 + q3
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Recognizing Perfect Cubes (X)3
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Why? You’ll do homework easier, score higher on tests.
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Memorize some common perfect cubes of integers
1
13
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8
23
27
33
64
43
125
53
216 … 1000
63 … 103
Unlike squares, perfect cubes of negative integers are different:
-216 … -1000
(-6)3 … (-10)3
-1
-8 -27 -64 -125
(-1)3 (-2)3 (-3)3 (-4)3 (-5)3
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Flashback: Do you remember how to tell if an integer divides evenly by 3?
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Variables with exponents divisible by 3 are also perfect cubes
x3 = (x)3 y6 = (y2)3 -b15 = (-b5)3
Monomials, too, if all factors are also perfect cubes
a3b15 = (ab5)3 -64x18 = (-4x6)3 125x6y3z51 = (5x2yz17)3
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The Difference between 2 Cubes
X3 – Y3 = (X – Y)(X2 + XY + Y2)
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F3 – L3 factors easily to (F – L)(F2 + FL +L2)
Examine 27a3 – 64b3
(3a)3 – (4b)3
(3a – 4b)(9a2 + 12ab + 16b2)
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Remember to remove common factors and to factor completely
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p3 – 8
(p)3 – (2)3
2x6 – 128 = 2[x6 – 64]
2[(x2)3 – 43]
(p – 2)(p2 + 2p + 4)
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2(x2 – 4)(x4 + 4x2 + 16)
2(x + 2)(x – 2)(x4 + 4x2 + 16)
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The Sum of 2 Cubes
X3 + Y3 = (X + Y)(X2 – XY + Y2)
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F3 + L3 factors easily to (F + L)(F2 – FL +L2)
Examine 27a3 + 64b3
(3a)3 + (4b)3
(3a + 4b)(9a2 – 12ab + 16b2)
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Remember to remove common factors and to factor completely
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p3 + 8
(p)3 + (2)3
2x6 + 128 = 2[x6 + 64]
2[(x2)3 + 43]
(p + 2)(p2 – 2p + 4)
2(x2 + 4)(x4 – 4x2 + 16)
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Perfect x3 – y3 = (x – y)(x2 + xy + y2)
Cubes x3 + y3 = (x + y)(x2 – xy + y2)
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p3 + q3 = (p + q)( p2 – pq + q2)
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216x3–1000 = (6x)3–(10)3 = (6x–10)(36x2+60x+100)
= 8(27x3–125) = 8((3x)3–(5)3) = 8(3x-5)(9x2+15x+25)
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27a3 – 11 = prime (middle term can’t disappear unless both are 3 )
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x6 – 64 = (x2)3–(4)3=(x2–4)(x4+4x2+16)= (x+2)(x-2)(x4+4x2+16)
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(p + q)3 + r3 = (p + q + r)((p+q)2 – (p+q)r + r2)
= (p + q + r)(p2 + 2pq + q2 – pr – qr + r2)
Ready for Your Homework?
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5.6
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What Next?
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Section 5.7 General
Factoring Strategy
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Look for patterns …
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