I'm having a little trouble with problems that have an unspecified polynomial, for example $p(x)$, and having to get properties of them. A problem I ran across had something along the lines of $p(x) \div (x-25)=0$, then asked to find for which values of $p(x)$ make this statement true. Examples include $p(25)=0$, etc. I know that $p(x)$ is not 25, because dividing by zero is impossible.

Any help would be much appreciated!

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    $\begingroup$ You don’t mean ambiguous: you mean unspecified. $\endgroup$ – Brian M. Scott Oct 27 '13 at 2:14
  • $\begingroup$ Sorry, thanks for the correction. $\endgroup$ – Carpetfizz Oct 27 '13 at 2:15
  • $\begingroup$ I think you have stated the conditions erroneously. ${p(x)\over x-25}=0$ can only have solutions where $p(x)=0$, and probably also $x\ne 25$ although that depends somewhat on the degree of $p(x)$. $\endgroup$ – abiessu Oct 27 '13 at 2:17
  • $\begingroup$ That's where I got confused. The question I wrote is how it appeared. It was multiple choice, so it had options such as $p(-25)=0$, $p(-5)=0$, etc. $\endgroup$ – Carpetfizz Oct 27 '13 at 2:18
  • $\begingroup$ What is the definition of your $\div$? I suspect $f(x) \div g(x)$ is either the quotient $q(x)$ or the remainder $r(x)$ when you perform a long division of the polynomial $f(x) = g(x)q(x)+r(x)$ by polynomial $g(x)$. $\endgroup$ – achille hui Oct 27 '13 at 2:24

The statement $p(x) \div (x-25) = 0$ is ambiguous, there are 3 possible interpretations of it.

  1. ordinary division as a rational function $\frac{p(x)}{x-25} = 0$.
  2. the quotient $q(x)$ in a long division is zero.
  3. the remainder $r(x)$ in a long division is zero.

In the $2^{nd}$ and $3^{rd}$ interpretation, the function $q(x)$ and $r(x)$ are uniquely determined by the expansion of $$p(x)\quad\text{ as }\quad q(x)(x - 5) + r(x)$$ where $q(x)$ and $r(x)$ are polynomials. Furthermore, $\deg r(x) < \deg(x-5) = 1$ forces $r(x)$ to be a constant.

In the original multiple-choice question, it asks which one of the four values $p(\pm 5)$, $p(\pm 25)$ is zero.

  1. In the $1^{st}$ interpretation, $p(x) = 0$ identically. This means $p(\pm 5) = p(\pm 25) = 0$. So the MC doesn't have an unique solution.

  2. In the $2^{nd}$ interpretation, $q(x) = 0$ implies $p(x) = r(x)$. Since $r(x)$ is a constant, either all or none of $p(\pm 5), p(\pm 25) = 0$. Once again, the MC doesn't have an unique solution.

  3. In the $3^{rd}$ interpretation, $r(x) = 0$ implies $p(x)$ can be factorized as $q(x)(x-25)$. From this, we get $p(25) = q(25)(25 - 25) = 0$. We don't have any information about what happens at $p(\pm 5)$ or $p(-25)$.

Combine these observations, it is a safe bet the $\div$ in the original question means taking the remainder and the answer of the MC is $p(25) = 0$.

  • $\begingroup$ Awesome, I think I understand it well now. Thank you so much! $\endgroup$ – Carpetfizz Oct 27 '13 at 3:21

Hint: the only way to get $A/B = 0$ is $A=0$.


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