Where do these $x-a, x-b,x-c$ come from and how?

Question

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

Answer 1

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 $Q_1,Q_2,Q_3$.

Answer 2

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 $R(x)=P(a)$