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I was reading about polynomials and long division at wikipedia, and came over this part.

Polynomial long division can be used to find the equation of the line that is tangent to the graph of the function defined by the polynomial $P(x)$ at a particular point $x = r$.

If $P(x)$ divided by $(x - r)^2$ leaves a remainder of $R(x)$ then the equation of the tangent line to $P(x)$ at $x = r$ is $y = R(x)$ (regardless of whether or not $r$ is a root of the polynomial).


Can someone give me an intuitive explenation of why this is true, or a proof? Googling gave me nothing, and a few drawings of using $(x-a)(x-\Delta x)$ seemed to have something to do with the line from $a$ to $b$. Now letting $\delta x \to a$ would give the tangent. Is this correct?

I do not quite see why dividing $P(x):(x-a)(x-\Delta x)$ would give me a line through $a$ and $\Delta x$ though..

Any help, suggestions or tips is very welcome =)

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Are you familiar with calculus? – cderwin Jan 6 '13 at 14:47
Undergrad in mathematics, so yeah I know some calculus ;) Was refreshing my knowledge for teaching some students later this year. – N3buchadnezzar Jan 6 '13 at 14:50
up vote 1 down vote accepted

In general, if $f$ and $g$ are functions, $f$ is tangent to $g$ at $a$ if $f(a)=g(a)$ and $f'(a)=g'(a)$.

We have $p(x)=(x-r)^2+r(x)$, which gives $$p'(r)=2(r-r)+r'(x)=r'(x)$$

So $p'(x)=r'(x)$

We also know that $p(r)=(r-r)^2+r(x)=r(x)$, so both stipulations for $p$ and $r$ to be tangent at $r$ are true.

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You're missing a $q(x)$ in $p(x)=(x-r)^2 + r(x)$. And you want to say $p'(r) = r'(r)$ and $p(r)=r(r)$ (which are weird since you chose $r$ to represent two different things). – Quinn Culver Sep 19 '14 at 12:27

Since you're dividing by a quadratic, the remainder is linear --- $R(x)=ax+b$ for some $a,b$. You have $$P(x)=(x-r)^2Q(x)+R(x)$$ for some polynomial $Q$. Differentiate, set $x$ equal to $r$, and take it from there. Come back if you need more help.

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