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ALGEBRA ALGEBRA + x2 - X + + Х (ii) x2 - 7x + 12 = x - 3x - 4x + 12 (x+3)(x3 - 7x+6)x3-7x+6 = x (x - 3) – 4 (x-3) (x - 1) (x+3) x-1 = (x-3) (x - 4) (iii) 2x2 - 11x + 15 = 2x2 - 6x-5x + 15 Now, x-1)x3-7X+6 (x2 + x-6 x x² 2x (x-3) - 5 (x+3) = (x - 3) (2x - 5) x - 7x + 6 HCF = 1 as no term is common. 4. If the LCM and HCF of two quadratic polynomi- - 6x + 6 als are x - 7x+6 and (x-1) respectively, find the - 6x + 6 og polynomials. (1) (x - 3x + 2), (x2 + 2x + 3) (2) (x + 3x - 2), (x2 - 2x + 3) :: Second polynomial q (x) = x2 + x-6 (3) (x2 – 3x + 2), (x + 2x - 3) (4) (x + 3x + 2), (x2 + 2x + 3) SOLVED OBJECTIVE TYPE QUESTIONS Sol. (3) Putting X- 1 = 0 i.e. x = 1 in (x - 1) respec- 1. The LCM of x3 - 1, *4 + x2 + 1 and 4 - 5x2 + 4 is : tively. (1) (x - 1) (x + 1) (x - 2) Remainder = (+1)3 – 7 (1) + 6 = 1-7 + 6 = 0 (2) (x - 1) (x + 1) (x + 2) .: (x - 1) is a factor of expression *' – 7x + 6 (3) (x - 1) (x2 - 4) Now, x3 - 7x + 6 = x (x - 1) + x (x - 1)-6 (x - 1) (4) (x2 - 1) (x® – 4) (x + x + 1) (x + 1 - x) (x - 1) (x + x-6) Sol. (4) (i) x2 - 1 = (x - 1) (x + x + 1) = (x - 1) (x2 + 3x - 2x - 6] (ii) x4 + x + 1 = x2 + 2x2 + 1 - x? = (x - 1) (x (x + 3) - 2 (x + 3)] : (x + 1)2 - X in = (x - 1) (x - 2) (x + 3) - (x + 1 - x) (x + 1 + x) LCM = x - 7x+ 6 = (x - 1) (x-2) (x + 3) and their (iii)x4 - 5x + 4 = x4 - 4x2 - y2 + 4 HCF = (x - 1) = x (x2 - 4) - 1 (x2 - 4) :: (x - 1) is common in both. .:. First expression = (x - 1) (x - 2) = x2 - 3x + 2 = - = A1..2 ...
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