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Let $p(x)$ be any polynomial with real coefficients, and $d(x)$ a monic quadratic polynomial. By the division algorithm we may write $$p(x) = d(x) q(x) + r(x)$$ where $\deg r(x) < 2$. Here, $q(x)$ is the quotient and $r(x)$, the remainder, is at most linear.

Now if the divisor has two distinct real zeros -- that is, if $d(x) = (x-a)(x-b)$ for some $a \ne b$ -- then the quotient and remainder have a natural geometric interpretation:

  • $r(x)$ gives the line that passes through the points $(a,p(a))$ and $(b, p(b))$.
  • $q(x)$ gives the leading coefficient of the unique parabola that passes through the three points $(a, p(a))$, $(b, p(b))$, and $(x, p(x))$.

If the divisor has only one zero with multiplicity 2 -- that is, if $d(x) = (x-a)^2$ for some $a$ -- then there is a slightly different geometric interpretation:

  • $r(x)$ gives the line that passes through $(a, p(a))$ and is tangent to $p(x)$ at that point
  • $q(x)$ gives the leading coefficient of the unique parabola that passes through $(a, p(a))$ and $(x, p(x))$ and is tangent to $p(x)$ at $a$.

My question: Is there a geometric interpretation of the quotient and remainder in the case where $d(x)$ has no real roots?

(And before anybody says it in an answer: I know that if you pass to complex numbers then the first interpretation above holds. I'm looking for a geometric interpretation of what's happening in $\mathbb{R}^2$. Such an interpretation -- if one exists at all -- would presumably not say anything about the roots of $d(x)$, but it might conceivably make reference to the real part of those roots, for example. )

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    Show me a form of $d(x)$ with no real roots without using complex roots, as you say in the last part. Since if you use complex roots to define $d(x)$, then you are no longer in $\mathbb{R^2}$ but in the Argand plane. – 2017-02-12
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    @SchrodingersCat Obviously if $d(x)$ has no real roots then it has complex roots. But I'm wondering if there is a geometric interpretation of $q(x)$ and $r(x)$ that describes what can be seen in the graphs of those functions in $\mathbb(R)^2$. – 2017-02-12
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    You are mixing things up. If d is complex, you are no longer in the real plane, you are dealing with the complex Argand plane. – 2017-02-12
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    A polynomial with complex roots nevertheless can be considered as a function from $\mathbb{R} \to \mathbb{R}$ and has a graph in the real plane. You don't *have* to think about where its roots live. – 2017-02-12
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    @SchrodingersCat If I take you literally, then you would say that a question like "Find the minimum of $d(x) = x^2 + 2x + 5$" would be unanswerable, because $d(x)$ has complex roots, so you have to think of it in the complex plane, but in the complex plane the function has no minimum. – 2017-02-12

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