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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 5.61

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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 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 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 x 2 = (x) 2 y 6 = (y 3 ) 2 y 6 = (-y 3 ) 2 a 2 b 14 = (ab 7 ) 2 Monomials, too, if all factors are also perfect squares a 2 b 14 = (ab 7 ) 2 81x 8 = (9x 4 ) 2 225x 4 y 2 z 22 = ( 15x 2 yz 11 ) 2 a 2 b 14 = (-ab 7 ) 2 81x 8 = (-9x 4 ) 2 225x 4 y 2 z 22 = (- 15x 2 yz 11 ) 2 … don’t forget those opposites! 5.62

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The Difference between 2 Squares F 2 – L 2 factors easily to (F + L)(F – L) Examine 49x 2 – 16 (7x) 2 – (4) 2 (7x + 4)(7x – 4) Remember to remove common factors and to factor completely 4U: 64a 4 – 25b 2 x 4 – 1 2x 4 y – 32y (8a 2 ) 2 – (5b) 2 (x 2 ) – 1 2 2y(x 4 – 16) (8a 2 + 5b)(8a 2 – 5b) (x 2 + 1)(x 2 – 1) 2y(x 2 + 4)(x 2 – 4) (x 2 +1)(x+1)(x-1) 2y(x 2 +4)(x+2)(x-2) 5.63

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p 2 + q 2 = prime! ( sum of 2 “simple” squares is never factorable ) 256x 2 – 100 = (16x) 2 – (10) 2 = (16x + 10)(16x – 10) 256x 2 – 100 = 4(64x 2 – 25) = 4(8x + 5)(8x – 5) 16a 2 – 11 = prime ( middle term can’t disappear unless both are 2 ) x 2 – (y + z) 2 = (x + y + z)(x – y – z) ( note that –(y+z)=–y–z ) x 2 + 6x + 9 – z 2 = (x + 3) 2 – z 2 = (x + 3 + z)( x + 3 – z) 3a 4 – 3 = 3(a 4 –1) = 3(a 2 +1)(a 2 –1) = 3(a 2 +1)(a+1)(a–1) Ready for Perfect Cubes? Perfectly Square Practice 5.64

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A Disappearing Act p 2 – pq + q 2 x p + q p 2 q – pq 2 + q 3 p 3 – p 2 q + pq 2 so, the sum is p 3 + q 3 = p 3 + q 3 5.65

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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 8 27 64 125 216 … 1000 1 3 2 3 3 3 4 3 5 3 6 3 … 10 3 Unlike squares, perfect cubes of negative integers are different: -1 -8 -27 -64 -125 -216 … -1000 (-1) 3 (-2) 3 (-3) 3 (-4) 3 (-5) 3 (-6) 3 … (-10) 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 x 3 = (x) 3 y 6 = (y 2 ) 3 -b 15 = (-b 5 ) 3 Monomials, too, if all factors are also perfect cubes a 3 b 15 = (ab 5 ) 3 -64x 18 = (-4x 6 ) 3 125x 6 y 3 z 51 = ( 5x 2 yz 17 ) 3 5.66

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F 3 – L 3 factors easily to (F – L)(F 2 + FL +L 2 ) Examine 27a 3 – 64b 3 (3a) 3 – (4b) 3 (3a – 4b)(9a 2 + 12ab + 16b 2 ) Remember to remove common factors and to factor completely p 3 – 8 2x 6 – 128 = 2[x 6 – 64] (p) 3 – (2) 3 2[(x 2 ) 3 – 4 3 ] (p – 2)(p 2 + 2p + 4) 2(x 2 – 4)(x 4 + 4x 2 + 16) 2(x + 2)(x – 2)(x 4 + 4x 2 + 16) The Difference between 2 Cubes X 3 – Y 3 = (X – Y)(X 2 + XY + Y 2 ) 5.67

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F 3 + L 3 factors easily to (F + L)(F 2 – FL +L 2 ) Examine 27a 3 + 64b 3 (3a) 3 + (4b) 3 (3a + 4b)(9a 2 – 12ab + 16b 2 ) Remember to remove common factors and to factor completely p 3 + 8 2x 6 + 128 = 2[x 6 + 64] (p) 3 + (2) 3 2[(x 2 ) 3 + 4 3 ] (p + 2)(p 2 – 2p + 4) 2(x 2 + 4)(x 4 – 4x 2 + 16) The Sum of 2 Cubes X 3 + Y 3 = (X + Y)(X 2 – XY + Y 2 ) 5.68

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Perfect x 3 – y 3 = (x – y)(x 2 + xy + y 2 ) Cubes x 3 + y 3 = (x + y)(x 2 – xy + y 2 ) p 3 + q 3 = (p + q)( p 2 – pq + q 2 ) 216x 3 –1000 = (6x) 3 –(10) 3 = (6x–10)(36x 2 +60x+100) = 8(27x 3 –125) = 8((3x) 3 –(5) 3 ) = 8(3x-5)(9x 2 +15x+25) 27a 3 – 11 = prime ( middle term can’t disappear unless both are 3 ) x 6 – 64 = (x 2 ) 3 –(4) 3 =(x 2 –4)(x 4 +4x 2 +16)= (x+2)(x-2)(x 4 +4x 2 +16) (p + q) 3 + r 3 = (p + q + r)((p+q) 2 – (p+q)r + r 2 ) = (p + q + r)(p 2 + 2pq + q 2 – pr – qr + r 2 ) Ready for Your Homework? 5.69

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What Next? Section 5.7 General Factoring Strategy Section 5.7 Look for patterns … 5.610

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