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Let $p$ be a real polynomial ($p\in\mathbb{R}[x])$, if

  • 5 is the remainder of the division $\frac{p(x)}{x-2}$, and
  • 2 is the remainder of the division $\frac{p(x)}{x-5}$

What is the reaminder of the division $\frac{p(x)}{(x-2)(x-5)}$?

I have tried it, but I have no idea... Thank you

Matemáticas
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    Write $,p(x) = (x-2)(x-5)q(x)+ax+b,$ then plug in a couple of well chosen values for $x$. – dxiv Feb 04 '18 at 06:51

1 Answers1

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$p(x)=(x−2)(x−5)q(x)+ax+b$, for some $a,b \in \mathbb{R}$

Since we know the remainder of $p(x)$ when divided by $x-2$ and $x-5$

we obtain the following \begin{align} p(2) &= (2-2)(2-5)q(2) + a(2)+b\\ &= 2a+b = 5 \label{a} \end{align}

\begin{align} p(5) &= (5-2)(5-5)q(5) + a(5)+b\\ &= 5a+b = 2 \label{b} \end{align}

Solving the simultaneous equations \ref{a} and \ref{b}, we obtain $a = -1 \text{ and } b = 7$.

Hence, the remainder of the division $\frac{p(x)}{(x−2)(x−5)}$ is $-x + 7$.

David
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