# Thread: Find the polynomials p (x)

1. ## Find the polynomials p (x)

Find all polynomials p(x) where xp(x -1) = (x-5)p(x) and p(6) = 5 factorials

2. Originally Posted by rcs
Find all polynomials p(x) where xp(x -1) = (x-5)p(x) and p(6) = 5 factorials
Hint :

If $p(x)$ satisfies $xp(x-1)=(x-5)p(x)$ then,

For $x=0$ , $0=-5p(0)$ that is, $p(0)=0$ .

For $x=1$ , $1p(0)=-4p(1)$ that is, $p(1)=0$ .

...

$x=0,1,2,3,4$ are roots of $p(x)$

Fernando Revilla

3. 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).

4. Originally Posted by HallsofIvy
Didn't we just recently have this?
Yes.

5. ## answer from the book without solution

hello everybody Dear Mathematicians... i saw the answer from where i got that problem i posted above... this is the answer but i don't know how they reach to this

6. Originally Posted by rcs

hello everybody Dear Mathematicians... i saw the answer from where i got that problem i posted above... this is the answer but i don't know how they reach to this
We have proved that $x=0,1,2,3,4$ are roots of $p(x)$ so, all the polynomials satisfying

$xp(x-1)=(x-5)p(x)\quad[*]$

have the form:

$p(x)=x(x-1)(x-2)(x-3)(x-4)q(x)$

On the other hand, $[*]$ implies:

$x(x-1)(x-2)(x-3)(x-4)(x-5)q(x-1)=$

$(x-5)x(x-1)(x-2)(x-3)(x-4)q(x)$

that is, $q(x)=q(x-1)$ and necessarily $q(x)=a$ constant ( Why ?).

So, all polynomials $p(x)$ satisfying $[*]$ are:

$p(x)=ax(x-1)(x-2)(x-3)(x-4)$

and the condition $p(6)=5!$ implies $a=1/6$ .

Fernando Revilla