Math, asked by akangkshya, 1 year ago

factorise: x^3-2x^2-x+2


julianviera45: {x}^{3} - 2 {x}^{2} - x + 2 \\ = {x}^{2} (x - 2) - (x - 2) \\ = ( {x}^{2} - 1)(x - 2) \\ = ( {x}^{2} - {1}^{2} )(x - 2) \\ = (x - 1)(x + 1)(x - 2) \\ \\ hence \: factorised \:
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Answers

Answered by Anonymous
2
Let take f(x) = x3 - 2x2 - x + 2
The constant term in f(x) is are ±1 and ±2 Putting x = 1 in f(x), we have f(1) = (1)3 - 2(1)2 -1 + 2 = 1 - 2 - 1 + 2 = 0
According to remainder theorem f(1) = 0 so that (x - 1) is a factor of x3 - 2x2 - x + 2 Putting x = - 1 in f(x), we have f(-1) = (-1)3 - 2(-1)2 –(-1) + 2 = -1 - 2 + 1 + 2 = 0
According to remainder theorem f(-1) = 0 so that (x + 1) is a factor of x3 - 2x2 - x + 2 Putting x = 2 in f(x), we have f(2) = (2)3 - 2(2)2 –(2) + 2 = 8 -82 - 2 + 2 = 0
According to remainder theorem f(2) = 0 so that (x – 2 ) is a factor of x3 - 2x2 - x + 2 Here maximum power of x is 3 so that its can have maximum 3 factors

julianviera45: {x}^{3} - 2 {x}^{2} - x + 2 \\ = {x}^{2} (x - 2) - (x - 2) \\ = ( {x}^{2} - 1)(x - 2) \\ = ( {x}^{2} - {1}^{2} )(x - 2) \\ = (x - 1)(x + 1)(x - 2) \\ \\ hence \: factorised \:
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