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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. Recognizing Perfect Squares (X) 2.
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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
Recognizing Perfect Squares (X)2 • Why? Because it enable efficient factoring! • Memorize the first 16 perfect squares of integers1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 25612 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 squaresx2= (x)2 y6= (y3)2y6= (-y3)2a2 b14 = (ab7)2 • Monomials, too, if all factors are also perfect squaresa2 b14 = (ab7)2 81x8 = (9x4)2 225x4y2z22= (15x2yz11)2 • a2 b14 = (-ab7)2 81x8= (-9x4)2 225x4y2z22= (-15x2yz11)2… don’t forget those opposites! 5.6
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 x4 – 1 2x4y – 32y(8a2)2 – (5b)2 (x2) – 122y(x4 – 16)(8a2 + 5b)(8a2 – 5b)(x2 + 1)(x2 – 1) 2y(x2 + 4)(x2 – 4)(x2+1)(x+1)(x-1) 2y(x2+4)(x+2)(x-2) 5.6
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
A Disappearing Act • p2 – pq + q2 • xp + q • p2q– pq2 + q3 • p3 – p2q + pq2 so, the sum is • p3 + q3 = p3 + q3 5.6
Recognizing Perfect Cubes (X)3 • Why? You’ll do homework easier, score higher on tests. • Memorize some common perfect cubes of integers1 8 27 64 125 216 … 1000 13 23 33 43 53 63 … 103 • 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 cubesx3= (x)3y6= (y2)3-b15= (-b5)3 • Monomials, too, if all factors are also perfect cubesa3b15 = (ab5)3 -64x18 = (-4x6)3 125x6y3z51 = (5x2yz17)3 5.6
The Difference between 2 CubesX3 – 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 factorsand to factor completely • p3 – 8 2x6 – 128 = 2[x6 – 64](p)3 – (2)3 2[(x2)3 – 43](p– 2)(p2 + 2p + 4) 2(x2 – 4)(x4 + 4x2 + 16) 2(x + 2)(x – 2)(x4 + 4x2 + 16) 5.6
The Sum of 2 CubesX3 + 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 factorsand to factor completely • p3 + 8 2x6 + 128 = 2[x6 + 64](p)3 + (2)3 2[(x2)3 + 43](p+ 2)(p2 – 2p + 4) 2(x2 + 4)(x4 – 4x2 + 16) 5.6
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
What Next? • Section 5.7 General Factoring Strategy • Look for patterns … 5.6
