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In the polynomial remainder theorem I've noticed one thing that triggers me

$P(x) = Q(x)D(x) + R(x)$

Here we say that the divisor is $D(x) = x-a$ now to show that $P(a) = R(x)$, people simply put $x-a = 0$ which contradicts the fact that we cannot divide by zero.
Another way to justify this is that we just put the value of $x$ as $a$ everywhere. But this again is the same thing since we are converting the divisor into zero which doesn't make sense.

So is there any way to justify this or some other way to look at it so that it doesn't end up as dividing by zero?

Thank you for seeing my question.

Ottavio
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Tensor69
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  • Welcome to Math.SE! Please, use MathJax to write formulas in your questions. You just have to put dollar symbols between formulas, i.e. $x^2$ – Ottavio Dec 29 '20 at 05:29
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    No division is involved in the statement that $P(x)=Q(x)D(x)+R(x)$: there are a multiplication and an addition, and the result is true no matter what value is substituted for $x$. – Brian M. Scott Dec 29 '20 at 05:29
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    Dividing $P(x)$ by $x-a$ and evaluating $P(x)$ at $a$ are not the same. The polynomial $x-a$ is non-zero, so the division of $P(x)$ by $x-a$ does not involve dividing by zero. The evaluation $x\to a$ that comes after the fact has no impact on the division that took place before it. – Randy Marsh Dec 29 '20 at 06:28
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    THANKS EVERYONE – Tensor69 Dec 29 '20 at 06:37
  • If you are happy with @Eric Towers answer and consider the issue resolved, you should mark his answer as the accepted answer. Green tick it. – Martin Hansen Dec 29 '20 at 09:03
  • The crux of the matter is that we do the "division" of polynomials before the evaluation. This can lead to some powerful results, e.g. see here for the case of such universal cancellation before evaluation, which yields a slick proof of Sylvester's determinant identity, and a purely algebraic definition of the derivative of a polynomial. These matters will be clarified if you study polynomial rings in a course in abstract algebra. – Bill Dubuque Dec 29 '20 at 19:25

1 Answers1

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It is called the division algorithm, but no division actually occurs. Instead we use addition and multiplication to demonstrate what would happen if the division were performed. So there is no explicit division by zero.

Additionally, the polynomial $x - a$ is not the zero polynomial, so even if there were division by $x-a$, it is not division by zero.

Suppose $a$ is a zero of $P(x)$, so $x-a$ is a factor of $P(x)$. That is, $P(x) = (x-a)S(x)$ where $S(x)$ has degree one less than $P$. We have: \begin{align*} (x-a)S(x) &= P(x) \\ &= Q(x)(x-a) + R(x) \text{.} \end{align*} This is an equality of polynomials, so the value on the left equals the value on the right for each choice of value of $x$. In particular, we may specialize $x$ to the value $a$. (There is still no division actually occurring here. Recall that we define division by "what we would have had to multiply (then add a remainder of bounded degree) to the divisor to obtain the dividend". We are literally following the definition of division to write down this multiplication and addition.) After this specialization of $x$, we obtain \begin{align*} (a-a)S(a) &= P(a) \\ &= Q(a)(a-a) + R(a) \text{.} \end{align*} Of course, $a-a = 0$ and multiplication by zero is very easy, so this simplifies to \begin{align*} 0 &= P(a) \\ &= 0 + R(a) \text{.} \end{align*} We discover that $P$ is zero at one of its zeroes, $a$, and the remainder, $R(a)$, on dividing $P$ by the linear polynomial corresponding to that zero, $x - a$, is zero at $a$ as well.

But this is not as strong as we want. We want the remainder on dividing by an exact divisor to be the zero polynomial, not just a polynomial that is zero at one place. That is, we want $R(x) = 0$ for all $x$, not just $R(a) = 0$. Let's go back. \begin{align*} (x-a)S(x) &= P(x) \\ &= Q(x)(x-a) + R(x) \text{.} \end{align*} We require that $R(x)$ has degree less than that of $D(x) = x-a$, which has degree $1$. So $R(x)$ has to have degree at most $0$, so $R(x)$ is constant. We know the value of $R$ at one point: $R(a) = 0$, so $R$ is the constantly zero function, $R(x) = 0$. This is what we wanted to happen.

Eric Towers
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