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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 5.6 1 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 5.6 … don’t forget those opposites! 2 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 5.6 3 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? 5.6 4 A Disappearing Act p2 – pq + q2 x p+q p2q – pq2 + q3 p3 – p2q + pq2 so, the sum is p3 + q3 = p 3 + q3 5.6 5 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 5.6 6 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) 5.6 2(x2 – 4)(x4 + 4x2 + 16) 2(x + 2)(x – 2)(x4 + 4x2 + 16) 7 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) 5.6 8 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? 5.6 9 What Next? Section 5.7 General Factoring Strategy Look for patterns … 5.6 10