Proving $n+3 \mid 3n^3-11n+48$ I'm really stuck while I'm trying to prove this statement:
$\forall n \in \mathbb{N},\quad (n+3) \mid  (3n^3-11n+48)$.
I couldn't even how to start.
 A: Assuming that you’re actually trying to prove that $n+3$ divides $3n^3-11n+48$, you can always simply do a polynomial long division:
$$\begin{array}{rrr|rr}
&&&&&3n^2&-&9n&+&16\\ \hline
n&+&3&3n^3&&&-&11n&+&48\\
&&&3n^3&+&9n^2\\ \hline
&&&&&-9n^2&-&11n\\
&&&&&-9n^2&-&27n\\ \hline
&&&&&&&16n&+&48\\
&&&&&&&16n&+&48\\ \hline
\end{array}$$
There’s no remainder, so
$$3n^3-11n+48=(n+3)(2n^2-9n+16)\;.$$
A: To expand on caveman's answer:  Note that polynomial long division works well here:
Dividing $$3n^3 - 11n + 48 \tag{dividend}$$ by $\;(n + 3)\;\;\text{(divisor)},\;$ gives a quotient of $$3n^2 - 9n + 16\tag{quotient}$$
and leaves a remainder of $0$.
$$\text{That is, }\quad\quad\frac{3n^3 - 11n + 48}{n+3} = \;3n^2 -9n + 16 $$ $$ \iff \;\;(n+3)(3n^2 - 9n + 16) = 3n^3 - 11n + 48$$
The result of dividing $\,(3n^3 - 11n + 48)\,$ by $\,(n+ 3)\,$, where $n$ is ANY $n \in \mathbb N$, gives an integer quotient: $\;n^2 - 9n + 16$, with no remainder.
$$\text{Therefore, we have }\quad (n+3)\mid (3n^3 - 11n + 48) \iff (3n^3 = 11n + 48) \equiv 0 \pmod {n+3}$$
A: There are two routine ways to see this:


*

*Do the polynomial long division $(3n^3 -11n + 48) / (n+3)$ and check that the remainder is $0$.

*A polynomial $f(n)$ is divisible by $(n - a)$ if and only if $f(a) = 0$. So in this example, $f(n) = 3n^3 -11n + 48$. Plugging $a = -3$ into $f$ we get $f(-3) = 0$, which shows that $f(n)$ is divisible by $(n - (-3)) = (n + 3)$.
A: If you wrote it backwards by accident then the proof is by
$$(3n^2 - 9n + 16)(n+3) = 3n^3 - 11n + 48$$

if you meant what you wrote then $n=0\;$ gives a counter examples since $48$ doesn't divide $3$.
A: Hint $3*(-3)^3-11*(-3)+48=-81+33+48=0$.
