# 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.

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The statement is false as written: for $n=1$, for instance, it’s not true that $40\mid 4$. I suspect that you mean $$n+3\mid 3n^3-11n+48\;.$$ $m\mid n$ means that $m$ divides $n$, i.e., that $n=km$ for some integer $k$. – Brian M. Scott Mar 26 '13 at 23:17
Perhaps the OP meant to write does not divide (vs. divides). Then it is true (and less trivial than the alternative reversed interpretation). – Math Gems Mar 26 '13 at 23:24

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

$$3n^3-11n+48=(n+3)(2n^2-9n+16)\;.$$

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+1 simply for having the patience to mathjax that calculation – Alexander Gruber Mar 27 '13 at 1:28

There are two routine ways to see this:

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

2. 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 $(n - (-3)) = (n + 3)$.

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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.

$$\text{Therefore, we have }\quad (n+3)\mid (3n^3 - 11n + 48) \iff (3n^3 = 11n + 48) \equiv 0 \pmod {n+3}$$

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Hint $3*(-3)^3-11*(-3)+48=-81+33+48=0$.

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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 $3$.

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@caveman I mean | – pourjour Mar 26 '13 at 23:14