polynomial question (out of practice)

Question

Find all polynomials $p(x)$ such that $(x+3)p(x) = x p(x+1)$ for all real x.

Ok, I am out of practice with this stuff. Here is what I have tried:

making $x = -3$ and making $x = -1$ does not help because I just go in circles. How do you know what numbers to plug in?

Start with the constant term of $p(x)$. What must it be in order for your condition to be satisfied? — hardmath, Oct 05 '13 at 19:18

njguliyev · Answer 1 · 2013-10-05T19:28:35.277

2

Hint: Put $x=0$, $x=-1$ and $x=-2$ to obtain $p(0)=p(-1)=p(-2)=0$. Now subsitute $p(x)=x(x+1)(x+2)q(x)$.

Then you get $q(x)=q(x+1)$ for all $x \in \mathbb{R}$.

edited Oct 05 '13 at 19:28

answered Oct 05 '13 at 19:23

njguliyev

Ok, thanks. So I did that, hopefully correctly, and I get 0-0 then -2=0, then -2=-2. – user98983 Oct 06 '13 at 12:25

1 Answers1