Transcript 5.6
Section 5.6 Review Difference of Two Squares
Sum & Difference of Two Cubes
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
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
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
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!
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The Difference between 2 Squares
F2 – L2 factors easily to (F + L)(F – L)
Examine 49x2 – 16
(7x)2 – (4)2
(7x + 4)(7x – 4)
Remember to remove common factors and to factor completely
4U:
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
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
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
Why? You’ll do homework easier, score higher on tests.
Memorize some common perfect cubes of integers
1
13
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
Flashback: Do you remember how to tell if an integer divides evenly by 3?
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)
F3 – L3 factors easily to (F – L)(F2 + FL +L2)
Examine 27a3 – 64b3
(3a)3 – (4b)3
(3a – 4b)(9a2 + 12ab + 16b2)
Remember to remove common factors and to factor completely
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)
F3 + L3 factors easily to (F + L)(F2 – FL +L2)
Examine 27a3 + 64b3
(3a)3 + (4b)3
(3a + 4b)(9a2 – 12ab + 16b2)
Remember to remove common factors and to factor completely
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)
p3 + q3 = (p + q)( p2 – pq + q2)
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)
27a3 – 11 = prime (middle term can’t disappear unless both are 3 )
x6 – 64 = (x2)3–(4)3=(x2–4)(x4+4x2+16)= (x+2)(x-2)(x4+4x2+16)
(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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What Next?
Section 5.7 General
Factoring Strategy
Look for patterns …
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