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Let $p(x)$ be a polynomial of 3rd degree.

We know that the division of $p(x)$ by $x-4$ gives us a remainder of 2 and divided by $x+2$ gives us the remainder of 1.

What's the remainder of $p(x)$ by $(x-4)(x+1)$?

I've used the remainder theorem but I don't seem to get anywhere...

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    $\begingroup$ By $(x-4)(x+1)$ or $(x-4)(x+2)$? $\endgroup$ – Bernard Nov 14 '15 at 20:44
  • $\begingroup$ It's really $(x-4)(x+1)$, I mean, that's what is written.. $\endgroup$ – Concept7 Nov 14 '15 at 20:46
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We can write $$p(x)=(x-4)(ax^2+bx+c)+2.$$

Since $p(-2)=1$, we have $$1=(-6)(4a-2b+c)+2\quad\Rightarrow\quad c=-4a+2b+\frac 16.$$

So, we can write $$p(x)=ax^3+(b-4a)x^2+\left(-4a-2b+\frac 16\right)x+16a-8b+\frac 43.$$

Thus, we have $$p(x)=(x-4)(x+1)(ax+b-a)+\color{red}{\left(-3a+b+\frac 16\right)x+12a-4b+\frac 43}.$$

The answer is the red part. Here, we cannot eliminate $a,b$.

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