chapter polynomials all formula tell
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Algebraic Identities (a + b)(a - b) = a2 - b2 (a + b)2 = a2 + b2 + 2ac (a - b)2 = a2 + b2 - 2ac (a + b)3 = a3 + b3 + 3a2b + 3ab2 (a - b)3 = a3 - b3 - 3a2b + 3ab2 Example 1: Simplify (3u + 5w)(3u – 5w) Using the algebraic identities (a + b)(a - b) = a2 - b2, we substitute a for 3u and b for 5w. (3u + 5w)(3u – 5w) = (3u)2 – (5w)2 = 9u2 – 25w2 Thus (3u + 5w)(3u – 5w) = 9u2 – 25w2 Example 2 : Using the algebraic identities to simplify (3a + 7b)2 Using (a+b)2 = a2+2ab+b2 In this case we need to substitute 3a for a as well as 7b for b (3a + 7b)2 = (3a)2 + 2(3a)(7b) + (7b)2 = 9a2+ 42ab + 49b2 Thus (3a + 7b)2 = 9a2+ 42ab + 49b2 Example 3: Simplify (5a – 7b)2 Using (a-b)2 = a2-2ab+b2 we have: (5a – 7b)2 = (5a)2 – 2(5a) (7b) + (7b)2 = 25a2 – 70ab + 49b2. Thus (5a – 7b)2 = 25a2 – 70ab + 49b2 Example 4 : Expand (2x + 1)3 Using identity (a + b)3 = a3 + b3 + 3a2b + 3ab2 we have: (2x + 1)3 = (2x)3 + (1)3+ 3(2x)(1)(2x + 1) = 8x3 + 1 + 6x(2x + 1) = 8x3 + 12x2 + 6x + 1 Example 5: Expand (2x - 3y)3. Using identity (a - b)3 = a3 - b3 - 3a2b + 3ab2 we have: (2x - 3y)3 = (2x)3 - (3y)3 - 3(haha2x)(3y)(2x - 3y) = 8x3 - 27y3 - 18xy(2x - 3y) = 8x3 - 27y3 - 36x2y + 54xy2 Example 6: If the values of a+b and ab are 4 and 1 respectively, find the value of a3 + b3. a3+ b3 = (a+b)3 – 3ab(a+b) = (4)3 – 3(1)(4) = 64 – 12 = 52 Example 7: Factorize 27x3 + y3 + z3 - 9xyz 27x3 + y3 + z3 - 9xyz = (3x)3 + (y)3 + (z)3 - 3(3x)(y)(z) = (3x + y + z){(3x)2 + (y)2 + (z)2 - (3x)(y) - (y)(z) - (z)(3x)} Using identity x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx) = (3x + y + z)(9x2 + y2 + z2 - 3xy - yz - 3zx)