![(GBR 1)](/media/m/c/f/c/cfc5975c4dd2a4db46a064e976e42757.png)
The polynomial
![P(x) = a_0x^k + a_1x^{k-1} + \cdots + a_k](/media/m/6/f/0/6f0816288943c99733af0e396974050a.png)
, where
![a_0,\cdots, a_k](/media/m/e/5/8/e58f7723e60906d718ca1968056a8b36.png)
are integers, is said to be divisible by an integer
![m](/media/m/1/3/6/1361d4850444c055a8a322281f279b39.png)
if
![P(x)](/media/m/c/d/7/cd7664875343d44cd5f96a566b582b0e.png)
is a multiple of
![m](/media/m/1/3/6/1361d4850444c055a8a322281f279b39.png)
for every integral value of
![x](/media/m/f/1/8/f185adeed9bd346bc960bca0147d7aae.png)
. Show that if
![P(x)](/media/m/c/d/7/cd7664875343d44cd5f96a566b582b0e.png)
is divisible by
![m](/media/m/1/3/6/1361d4850444c055a8a322281f279b39.png)
, then
![a_0 \cdot k!](/media/m/5/2/3/5234c908ae37bfbfe174bd9b15b791fd.png)
is a multiple of
![m](/media/m/1/3/6/1361d4850444c055a8a322281f279b39.png)
. Also prove that if
![a, k,m](/media/m/e/5/8/e581b53ca27eb9dc168132637821ff41.png)
are positive integers such that
![ak!](/media/m/1/4/2/14263533328076464beef57c33822049.png)
is a multiple of
![m](/media/m/1/3/6/1361d4850444c055a8a322281f279b39.png)
, then a polynomial
![P(x)](/media/m/c/d/7/cd7664875343d44cd5f96a566b582b0e.png)
with leading term
![ax^k](/media/m/2/7/9/2799fe2876d3231319b51b1d81ba3695.png)
can be found that is divisible by
%V0
$(GBR 1)$ The polynomial $P(x) = a_0x^k + a_1x^{k-1} + \cdots + a_k$, where $a_0,\cdots, a_k$ are integers, is said to be divisible by an integer $m$ if $P(x)$ is a multiple of $m$ for every integral value of $x$. Show that if $P(x)$ is divisible by $m$, then $a_0 \cdot k!$ is a multiple of $m$. Also prove that if $a, k,m$ are positive integers such that $ak!$ is a multiple of $m$, then a polynomial $P(x)$ with leading term $ax^k$can be found that is divisible by $m.$