Factorise $\displaystyle P(x)=x^3-7x-6$ I know that the constant term for this equation is -6 . Then , if (x-a) were to be a factor of P(x) , then a must be a factor of -6 . Why is it so ? My book doesn't further explain .
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Originally Posted by mathaddict Factorise $\displaystyle P(x)=x^3-7x-6$ I know that the constant term for this equation is -6 . Then , if (x-a) were to be a factor of P(x) , then a must be a factor of -6 . Why is it so ? My book doesn't further explain . Because when you put it in the form: $\displaystyle P(x) = (x-a)(x^2+bx+c) $ Then a and c must multiply to give -6, and hence both must be factors.
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