please help me on this. thank you so much
Find all polynomials p(x) where xp(x -1) = (x-5)p(x) and p(6) = 5 factorials
Hint :
If $\displaystyle p(x)$ satisfies $\displaystyle xp(x-1)=(x-5)p(x)$ then,
For $\displaystyle x=0$ , $\displaystyle 0=-5p(0)$ that is, $\displaystyle p(0)=0$ .
For $\displaystyle x=1$ , $\displaystyle 1p(0)=-4p(1)$ that is, $\displaystyle p(1)=0$ .
...
$\displaystyle x=0,1,2,3,4$ are roots of $\displaystyle p(x)$
Fernando Revilla
Didn't we just recently have this?
Here's how I would do it (rather different from the way FernandoRevilla did it). Taking x= 5, 5p(4)= 0 so that p(4)= 0. Since p is a polynomial, that means that p has a factor of x- 4: we can write p(x)= (x- 4)q(x) for some polynomial q.
Putting that into the given equation, xp(x- 1)= x((x- 1)- 4)q(x-1)= x(x- 5)q(x-1)= (x- 5)(x- 4)q(x). Since this is true for all x, we can cancel the "x- 5" terms an see that q satisfies
xq(x- 1)= (x- 4)q(x).
Talking x= 4, we have 4q(3)= 0 so that q(3)= 0 and we can write q(x)= (x- 3)r(x) for some polynomial r.
Now xq(x-1)= x(x-1- 3)r(x-1)= (x- 4)(x- 3)r(x). Since this is true for all x, we cancel the "x-4" terms and have
xr(x-1)= (x- 3)r(x). Taking x= 3, we have 3r(2)= 0 so that r(x)= (x- 2)s(x) for some polynomial s.
Get the point? Keep doing that until it terminates (it will).
We have proved that $\displaystyle x=0,1,2,3,4$ are roots of $\displaystyle p(x)$ so, all the polynomials satisfying
$\displaystyle xp(x-1)=(x-5)p(x)\quad[*]$
have the form:
$\displaystyle p(x)=x(x-1)(x-2)(x-3)(x-4)q(x)$
On the other hand, $\displaystyle [*]$ implies:
$\displaystyle x(x-1)(x-2)(x-3)(x-4)(x-5)q(x-1)=$
$\displaystyle (x-5)x(x-1)(x-2)(x-3)(x-4)q(x)$
that is, $\displaystyle q(x)=q(x-1)$ and necessarily $\displaystyle q(x)=a$ constant ( Why ?).
So, all polynomials $\displaystyle p(x)$ satisfying $\displaystyle [*]$ are:
$\displaystyle p(x)=ax(x-1)(x-2)(x-3)(x-4)$
and the condition $\displaystyle p(6)=5!$ implies $\displaystyle a=1/6$ .
Fernando Revilla