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...

  • 2
    $\begingroup$ 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

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 $a,b$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.