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Suppose we have a polynomial $P(x)$

$$P(x) = x^3 - 8x^2+6x-k$$

and it is given that

$$P(a) = P(b) = P(c) = 3$$

I noticed that my teacher wrote down some equations such as

$$P(x) = \color{blue}{(x-a)}Q(x) +3$$

$$P(x) = \color{blue}{(x-b)}B(x) +3$$

$$P(x) = \color{blue}{(x-c)}C(x) +3$$

Where do these $x-a, x-b,x-c$ come from and how? I'll be glad if you explain.

Regards

Bobtrollsten
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2 Answers2

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Hint: Take into consideration that $P(\rho)$ is equal to the remainder of the division of $P(x)$ with $(x-\rho)$.

Edit: Consider that, dividing $P(x)$ with $(x-a)$ yields: $$P(x)=(x-a)A(x)+r,$$ where $r\in\mathbb{R}$ is the remainder. Since $P(a)=3$, we have that: $$P(a)=(a-a)A(a)+r\Leftrightarrow3=r,$$ and, as a result: $$P(x)=(x-a)A(x)+3.$$

Also, as pointed out by @Henning Makholm, there should be three different polynomials $Q_1,Q_2,Q_3$.

Vassilis Markos
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    It still seems to be unclear. – Bobtrollsten Mar 04 '19 at 17:47
  • I've added some small proof to help you! :) Tell me about any problems. – Vassilis Markos Mar 04 '19 at 17:52
  • What if we consider $P(x^2 + 1)$ instead of $P(x)$? For instance, can we still say that $P(x^2 + 1) = (x^2 + 1 -a)Q(x) + 3$? – Bobtrollsten Mar 04 '19 at 17:59
  • Then, $P(x^2+1)$ can be writen in the form $P(x^2+1)=(x-a)Q(x)+r$, where $r$ is the remainder. Actually, letting $x=a$ in the previous equation yields $r=P(a^2+1)$. Could you provide some more information about the level of the class you are attending etc so as to know where to direct your focus to? :) – Vassilis Markos Mar 04 '19 at 18:02
  • I'm grade 10 student. – Bobtrollsten Mar 04 '19 at 18:02
  • Alright, so, judging by what your teacher mentioned, shall I asume that you have been taught about the Euclidean Algorithm and, in general, about division between polynomials? – Vassilis Markos Mar 04 '19 at 18:05
  • Yes, please. It does not matter. – Bobtrollsten Mar 04 '19 at 18:05
  • So, you can divide any polynomial $p(x)$ by a first degree polynomial $x-a$ and find two other polynomials $q(x)$ and $r$ (the constant polynomial) such as: $$p(x)=(x-a)q(x)+r.$$Excuse me for the late reply! – Vassilis Markos Mar 04 '19 at 18:45
  • Is this just for first degree polynomials? – Bobtrollsten Mar 04 '19 at 19:30
  • No, this is true for polynomials of any degree. More specifically, if $p(x),q(x)$ are two polynomials, then there exist two unique polynomials $\pi(x),r(x)$ such that: $$p(x)=\pi(x)q(x)+r(x),$$ with $r(x)$ being zero or having degree less than that of $q(x)$. – Vassilis Markos Mar 04 '19 at 19:37
  • Is it also important that their input is $x$? because it wouldn't be valid for $P(2x^2+2)$ or somewhat similar. – Bobtrollsten Mar 04 '19 at 20:08
  • $p(2x^2+2)$ is a polynomial of $x$, as well. You can name it $q(x)$, if it seems more comfortable to you. In general, $p(q(x))$ is a polynomial of $x$, if $p,q$ are polynomials. – Vassilis Markos Mar 04 '19 at 20:19
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Your teacher should have written something like

\begin{align} P(x)=(x-a)Q(x)+3 \\ P(x)=(x-b)R(x)+3 \\ P(x)=(x-c)S(x)+3 \\ \end{align} where the equalities result from the Euclidean division of $P(x)$ by $x-a,\:x-b$ and $x-c$ respectively.

The general formula for the division by these polynomials is $$P(x)=(x-a)Q(x) +R(x),\quad \deg R<\deg(x-a)=1,$$ so the remainder $R(x)$ is a constant. It is easy to check the remainder is simply $R(x)=P(a)$

Bernard
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