Quotient of two polynomials with integer coefficients Let $f(x)$ a polynomial with integer coefficients. If $r$ is an integral root of $f$, then prove that the polynomial $\frac{f(x)}{x-r}$ is also a polynomial with integer coefficients. 
 A: If $r$ is an integral root of $f(x)$, then the first degree, monic polynomial $x-r$ is a factor of $f(x)$. This is a consequence of the Euclidean division of polynomials: we can equivalently say that considering $f(x)$ as a divident and $x-r$ as a divisor, the quotient is some integer polynomial $q(x)$ and the remainder is $0$. Thus:
$$
f(x)=q(x)(x-r)
$$
Now the fact that $q(x)$ is again an integer polynomial, is a result of the fact that generally in Euclidean division of polynomials if the divident $\Delta(x)$ and the divisor $\delta(x)$ are both integer polynomials and furthermore the divisor is monic (i.e. has leading coefficient equal to $1$) then the quotient $q(x)$ and the remainder $r(x)$:
$$
\Delta(x)=\delta(x)q(x)+r(x)
$$
are again integer polynomials. This can be easily seen from the long-division scheme. In fact, it is easily seen (considering long division) that in such a situation, the leading coefficient of the quotient $q(x)$ will be equal to the leading coefficient of the divident $\Delta(x)$. 
P.S.: However, notice that if the divisor is not monic, then the quotient and the remainder are in general rational polynomials (even when the divident and the divisor are integer polynomials). Euclidean division of polynomials is generally defined inside $\mathbb{Q}[x]$. Only if the divisor is monic, can it  take place inside $\mathbb{Z}[x]$. 
A: Let $f(x)$ be a polynomial of order $n$ with integer coefficients $a_i$.
$$
f(x) =\sum_{i=0}^{n} a_i x^i
$$
As $r$ is a root of $f(x)$ then $x-r$ divides $f(x)$ with quotient $q(x)$ and zero remainder. $f(x)$ can be expressed as
$$f(x)=(x-r)q(x)+0$$
Let $q(x)$ be a polynomial of order $n-1$ with coefficients $b_i$.
$$
q(x)=\sum_{i=0}^{n-1} b_i x^i
$$
Combining the given expressions above, we have
\begin{align}
\sum_{i=0}^{n} a_i x^i
&= (x-r)\sum_{i=0}^{n-1} b_i x^i \\
&= x\sum_{i=0}^{n-1} b_i x^i -r\sum_{i=0}^{n-1} b_i x^i \\
&= \sum_{i=0}^{n-1} b_i x^{i+1} -\sum_{i=0}^{n-1} rb_i x^i \\
&= \sum_{i=1}^{n} b_{i-1} x^i -\sum_{i=0}^{n-1} rb_i x^i \\
&= \left(b_{n-1}x^n+\sum_{i=1}^{n-1} b_{i-1} x^i\right) -\left(\sum_{i=1}^{n-1} rb_i x^i+rb_0\right) \\
a_nx^n+\sum_{i=1}^{n-1}a_ix^i+a_0
&=b_{n-1}x^n+\sum_{i=1}^{n-1} \left(b_{i-1} -rb_i \right)x^i  - rb_0 
\end{align}
By equating the corresponding coefficients, we have

*

*As $b_{n-1}=a_n$ then $b_{n-1}$ is an integer.

*Recursively for $i\in\{n-1,\cdots,1\}$, the coefficients $b_{i-1}=a_i+rb_{i}$ are all integers.

Thus the coefficients of $q(x)$ are all integers. Done!
