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I have recently started learning about polynomials. I've been able to grasp polynomial long division algorithm and the remainder and factor theorems and also a few other common-sense theorems about polynomials. There's just one property of polynomials I don't quite understand the proof of.

The property: "If x - a is a factor of polynomial P(x), then a is a factor of the constant term of the polynomial."

There are 2 proofs that I've seen so far that prove this theorem. The first proof I understand and makes complete sense to me. In my view, I think proof 1 is easier to understand.

Proof 1: enter image description here

Proof 2:

enter image description here

The second proof is the one I don't understand. More specifically, the part that I don't understand is how:

enter image description here

Can someone please carefully explain how those two expressions are equal to each other? I just don't see how those expressions are equal. I can't find any common factors that have been taken out or what logic has been used to rewrite the expression in that way.

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  • $\begingroup$ What is the difficulty? You just collect terms on the rhs according to the power of $x$. Actually, all you care about is the constant term, which is obviously $aq_0$. Yes, there is obviously a serious error in the last long line: $q_{n-2}-aq_{n-1}$ has been written without the $q_{n-1}$ and similarly for the other terms. But the leading term and the constant term are correct. $\endgroup$ – almagest Jan 23 '20 at 13:49
  • $\begingroup$ So you're saying the last line of work in proof 2 is incorrect? It seems like they haven't factored properly, they missed some of the coefficients for the corresponding x's, but I wasn't sure if it was incorrect or I was just not seeing something. $\endgroup$ – Dom Turner Jan 23 '20 at 22:16
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I presume you're talking about polynomials with integer coefficients: if the coefficients can be real numbers, the question makes no sense because any nonzero real number can divide any other real number.

Then the point is that $p(x) = q(x) r(x)$ implies $p(0) = q(0) r(0)$, where $p(0)$, $q(0)$ and $r(0)$ are the constant terms of $p, q$ and $r$.

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  • $\begingroup$ Yes, your right. I forgot to mention that in the title but the proof does assume that you are dealing with a polynomial with integer coefficients and that the real zeroes of the polynomial are also integers. $\endgroup$ – Dom Turner Jan 24 '20 at 10:31
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This does not answer your very question, but the theorem seems easy.

If $(x-a)$ divides $P(x)$ then $P(a)=0$. But $P(a)=a(p_na^{n-1}+\cdots+a p_2+p_1)+p_0$ and $p_0=-a(\cdots)$.

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@DomTurner Just to amplify my comment under the question. You have $P(x)=p_nx^n+\dots+p_0$ and $Q(x)=q_{n-1}x^{n-1}+\dots+q_0$. So the 2nd proof should have $$(x-a)Q(x)=xQ(x)-aQ(x)=$$ $$q_{n-1}x^n+q_{n-2}x^{n-1}+\dots+q_1x^2+q_0x-a(q_{n-1}x^{n-1}+\dots+q_0)$$ $$=q_{n-1}x^n+(q_{n-2}-aq_{n-1})x^{n-1}+(q_{n-3}-aq_{n-2})x^{n-2}+\dots+(q_0-aq_1)x-aq_0$$ At that point there is a serious typo in your source because it missed out all those $q_i$ by putting $(q_i-a)x^{i+1}$ for each term instead of $(q_i-aq_{i+1})x^{i+1}$. But it doesn't affect the important point which is that the constant term is $aq_0$.

Of course, you are correct that when one is struggling to learn something, typos in a trusted source make things much more difficult.

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  • $\begingroup$ Oh ok, that clarified everything. I was wondering whether or not there was a typo in the second proof but I guess it doesn't matter because it doesn't affect the point of the proof which is to show that every integer zero of a polynomial is a factor of the constant term of the polynomial (assuming the polynomial has integer coefficients). Note that I was suspecting that there was an error in the second proof before submitting this question, I just wanted to verify my thought. Thanks. $\endgroup$ – Dom Turner Jan 24 '20 at 10:29

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