The remainders of dividing the polynomial $P(x)$ with $x-1$, $x-2$, $x+1$ is $2$, $3$, and $6$ respectively. What is the remainder of dividing the polynomial $P(x)$ with $(x-1)(x-2)(x+1)$?
So what I could gather from this:
$P(x) = (x-1)D_1(x)+2$
$P(x) = (x-2)D_2(x)+3$
$P(x) = (x+1)D_3(x)+6$
from which I know:
$P(1)=2$
$P(2)=3$
$P(3)=6$
What I know I should find is $R(x)$ in:
$P(x) = (x-1)(x-2)(x+1)D_4(x) + R(x)$
now where I'm not sure, but what I think is that $P(x)$ should be in the form of
$$P(x)=Ax^4+Bx^3+Cx^2+Dx+E$$
and from my previous statements of what $P(1)$, $P(2)$ and $P(3)$ are equal to, I can gather that:
$P(1): A+B+C+D+E = 2$
$P(2): 16A+8B+4C+2D+E=3$
$P(3): 81A+27B+9C+3D+E=6$
now I have 3 equations with 5 unknown values. Is there anything I am missing? My actual thought is that I shouldn't be trying to find $P(x)$ and instead go for finding the remainder directly somehow, but I do not know how should I start with that.
$\endgroup$3 Answers
$\begingroup$The remainder $R(x)$ of the division of $P(x)$ by $(x-1)(x-2)(x+1)$ has degree at most $2$. So, it is a polynomial of the type $Ax^2+Bx+C$. And you know that $R(1)=2$, that $R(2)=3$ and that $R(-1)=6$. Can you take it from here?
$\endgroup$ 1 $\begingroup$The remainder has to be, at most, one degree less than the divisor. Since the divisor is degree $3$, it follows that the remainder is at most degree $2$, and this can be written in the form $Ax^2+Bx+C$.
Define $R(x):=Ax^2+Bx+C$. We will use this to write $P(x)$ in terms of $A$, $B$, and $C$.
\begin{align} P(1)&=A+B+C=2\tag1\\ P(2)&=4A+2B+C=3\tag2\\ P(-1)&=A-B+C=6\tag3\\ \end{align}
The reduction of the matrix is$$\left.\left[\begin{array}{ccc|c} 1&1&1&2\\ 4&2&1&3\\ 1&-1&1&6 \end{array}\right]\longrightarrow\left[\begin{array}{ccc|c} 1&0&0&1\\ 0&1&0&-2\\ 0&0&1&3 \end{array}\right]\right..$$
Therefore, the remainder is $\boxed{R(x)=x^2-2x+3}$.
$\endgroup$ $\begingroup$Hint
WLOG $$P(x)=(x-1)(x-2)(x-3)Q(x)+a(x-1)(x-2)+b(x-1)(x+1)+c(x+1)(x-2)$$
$$x=1\implies c(1+2)(1-2)=P(1)=2$$
Can you take it from here?
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