Thread:

1. factor theorem

Factorise $P(x)=x^3-7x-6$

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 .

2. Originally Posted by mathaddict
Factorise $P(x)=x^3-7x-6$

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:

$P(x) = (x-a)(x^2+bx+c)$

Then a and c must multiply to give -6, and hence both must be factors.
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