I'm supposed to show that $x-1$ is a factor of a polynomial P of positive degree if and only if the sum of the coefficients of P is zero. How do I do that exactly?
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4If $x-1$ is a factor, then what does that tell us about $P(1)$? – Trevor Norton Aug 26 '18 at 21:54
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That the polynomial would be equal to zero. But what does it mean to be a positive degree? Does it mean that $x^n$ where n starts with a positive number like $x^2$, $x^4$ and such? – Math Love Aug 26 '18 at 21:58
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Yes, positive degree means it is a polynomial of degree at least $1$, so that degree $0$ (constant) polynomials are not considered. – Jair Taylor Aug 26 '18 at 21:59
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Got it! Thank you – Math Love Aug 26 '18 at 22:01
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If $x-1$ is a factor then you have $p(x) = (x-1)q(x)$ where q(x) $ is another polynomial.
Let $x=1$, we get $$p(1) = (1-1)q(1)=0$$
That is if you let $x=1$ in your polynomial you will get $p(1)=0$
Note that when you find $p(1)$ you let $x=1$ , so you are just adding the coefficients of $P(x)$ together.
Thus if $x=1$ is a factor, the sum of coefficients of $p(x)$ is $0$.
Mohammad Riazi-Kermani
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The remainder of the division of a polynomial $p(x)$ by $x-a$ is $p(a)$, so $$p(x)\;\text{ divisible by }\;x-a\iff p(a)=0.$$ On the other hand $p(1)$ is just the sum of the coefficients of $p(x)$, whence the conclusion.
Bernard
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