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My class has just been taught about polynomial division, and how it can be used to see if something is a factor (although remainder theorem is quicker), if the remainder = 0.

But what is the use of remainder that does not equal 0, I'm unsure where to apply this to in situations.

For example if you do:

$$\frac{3x^3+5x^2+x+4}{x-2}$$

the remainder is $50$, which means (more on the graphic side of things), that the line $x=2$ will intercept $3x^3+5x^2+x+4$ at $(2,50)$

But what use is this beyond "find the intercept", "find the remainder"?

(Like how you can ask "5 - 0.2 = ?", but a situation you apply that to, could be: "what is the change from a £5 note used to buy a 20p pencil")

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If the remainder from dividing $f(x)=3x^3+5x^2+x+4$ by $x-2$ is $50$, then $f(2)=50$. Synthetic division is useful for evaluating polynomials quickly; the method is often associated with the name of Horner. –  J. M. Nov 19 '11 at 18:27
    
@J.M. Oh! I had thought it was the other way round, that explains a lot :) –  Jonathan. Nov 19 '11 at 18:53
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@Jonathan How do you calculate something like $\int \frac{3x^3+5x^2+x+4}{x-2} dx$? –  N. S. Nov 19 '11 at 20:20

3 Answers 3

In addition to the use mentioned by J.M. in the comments, you can also use remainders of polynomial division to find the tangent line to a polynomial at a particular point.

Concretely, if the remainder of $P(x)$ when divided by $(x-a)^2$ is $R(x)$, then the equation of the tangent line to $P$ at $x=a$ is $y = R(x)$.

How does this work? Note that if the remainder when $P$ is divided by $(x-a)^2$ is $R$ then

$$P(x) = (x-a)^2 Q(x) + R(x)$$

where $R(x)$ is linear. If we evaluate $P$ at $a$ then we get

$$P(a) = (a-a)^2 Q(a) + R(a) = R(a)$$

so $R$ and $P$ have the same value at $x=a$. Similarly if we evaluate the derivative of $P$ at $a$ we get

$$P'(a) = 2(a-a)Q(a) + (a-a)^2Q'(a) + R'(a) = R'(a)$$

so $P$ and $R$ have the same slope. But $R$ is linear, so it is completely determined by knowing its slope and value at one point. Therefore $y=R(x)$ is the tangent line to $P$ at $x=a$.

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It will take me a while to understand your answer fully, but thanks anyway. Also surely using differentiation is an easier method to finding the tangent? –  Jonathan. Nov 19 '11 at 19:05
    
In short: polynomial division, Taylor ;) expansion of a polynomial, and polynomial evaluation are essentially the same thing, seen in different ways. –  J. M. Nov 19 '11 at 23:45

One important theoretical application of division with remainder is that it can be used to generalize Euclid's algorithm for greatest common divisors from integers to polynomials. Say that we have some monstrous rational expression that we want to simplify: $$ \frac{p(x)}{q(x)}=\frac{x^{12}-3 x^9-x^8+x^7+16 x^5-3 x^4-13 x+13}{x^6-3 x^5-x^2+4 x-3}$$ Then, in general the best we can do is to find a common divisor of $p$ and $q$ and divide that out of the fraction. We can do this by using Euclid's algorithm. Start by dividing $p/q$ with remainder. We find that $$ p(x) = (x^6+3x^5+9x^4+24x^3+72x^2+216x+648)q(x) + 1957x^5-1957x+1957$$ Now let $r(x)= 1957x^5-1957x+1957$ and continue dividing $q/r$: $$ q(x) = \left(\frac{1}{1957}x-\frac{3}{1957}\right)r(x) + 0$$ We have reached a remainder of 0, so the greatest common divisor is the previous remainder -- but for polynomials this is only a greatest-degree common; any real multiple of $r(x)$ is of course also a common divisor. So for simplicitly we can take $\frac1{1957}r(x)=x^5-x+1$ and divide through by that (in two last divisions that are guaranteed to have zero remainder) to get $$\frac{x^7-3 x^4+13}{x-3}$$

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When you divide $3x^3+5x^2+x+4$ by $x-2$, the quotient is $3x^2+11x+23$ and the remainder is $50$. It follows that $$ \frac{3x^3+5x^2+x+4}{x-2} = 3x^2+11x+23 + \frac{50}{x-2}. $$ This fraction clearly approaches $0$ as $x$ approaches $\pm\infty$, so they graph of $y=\dfrac{3x^3+5x^2+x+4}{x-2}$ approaches the graph of $y=3x^2+11x+23$, which is a parabola, as $x$ approaches $\pm\infty$. The fraction at the end is the part that tells you have the graph of $y=\dfrac{3x^3+5x^2+x+4}{x-2}$ behaves at the vertical asymptote at $x=2$: it goes up to $+\infty$ as $x$ approaches $2$ from above (or "from the right" if you like), since $50/(x-2)$ is positive when $x>2$, and for a similar reason, it goes down to $-\infty$ as $x$ approaches $2$ from the other direction.

If you were integrating the function along a path in the complex plane that winds once counterclockwise around the number $2$, the only term that would matter would be $50/(x-2)$. The integral of the other part would be $0$. The integral of $50/(x-2)$ would be $50$ times what it would be if you had $1$ there instead of $50$.

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