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Question
the factors of x3−7x+6 are
the factors of x3−7x+6 are
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Solution
The correct option is D (x−1)(x+3)(x−2)
x3−7x+6=x3−1−7x+7=(x−1)(x2+x+1)−7(x−1)=(x−1)(x2+x+1−7)=(x−1)(x2+x−6)=(x−1)[x2+3x−2x−6]=(x−1)[x(x+3)−2(x+3)]=(x−1)(x+3)(x−2)
(x−1)(x+3)(x−2)
x3−7x+6=x3−1−7x+7=(x−1)(x2+x+1)−7(x−1)=(x−1)(x2+x+1−7)=(x−1)(x2+x−6)=(x−1)[x2+3x−2x−6]=(x−1)[x(x+3)−2(x+3)]=(x−1)(x+3)(x−2)
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