How can I arrange this expression to get the form of another? I need to get this:
$$4(n+1)^2-(n+1)$$
Initially I have this:
$$4n^2-n+(8n+3)$$
Which I simplify to this:
$$4n^2+7n+3$$
And since it is a cuadratic I simplify it to this:
$$(n+1)(4n+3)$$
This is how far I got. I know that my expression has the same value as the original, but I need to reach the same form as well.
I have trouble with this kind of problem. I always get stuck when needed to make a expression take some form.
Yes, I need a way to arrange my expression to get that. But it would be much better if you  could help me explaining those "techniques" or "tactics" I seem to be clearly lacking instead of just a solution to this specific scenario. Were my steps correct so far? Or should I have stopped earlier and taken a different route?
 A: Your expression simplified to $4n^2+7n+3$. That part of the work was fine. The factoring that you did after that was correct, but unnecessary.
We want to show that $4n^2+7n+3$ is equal to $4(n+1)^2 -(n+1)$. I suggest working with this expression. So expand $(n+1)^2$. We get $n^2+2n+1$. It follows that
$$\begin{align} 4(n+1)^2-(n+1)&=4(n^2+2n+1)-(n+1)\\&=(4n^2+8n+4)-(n+1)\\&=4n^2+7n+3.\end{align}$$
A: First note that if you only need to show that they are equal, it's enough to evaluate both into the same form, for example:
$$\begin{aligned}4(n+1)^2-(n+1)&=4(n^2+2n+1)-n-1\\&=4n^2+8n+4-n-1\\&=4n^2+7n+3\end{aligned}$$
and
$$4n^2-n+(8n+3)=4n^2+7n+3.$$
Since the expanded forms are equal, the originals are, too. However, now you also have a way to get to the first from the second by just backtracking:
$$\begin{aligned}4n^2-n+(8n+3)&=4n^2+7n+3\\&=4n^2+8n+4-n-1\\&=4(n^2+2n+1)-n-1\\&=4(n+1)^2-(n+1)\end{aligned}$$
A: Here’s another way to look at it. You have the expression $4n^2-n+(8n+3)$, which simplifies to $$4n^2+7n+3,$$ and you want to rewrite it as a polynomial in $n+1$. So let’s introduce a new variable $m=n+1$ and rewrite the expression in terms of $m$. Now $n=m-1$, so we substitute this into your expression to get $$4(m-1)^2+7(m-1)+3$$ $$= 4(m^2-2m+1)+7m-7+3$$ $$=4m^2-8m+4+7m-4$$ $$=4m^2-m$$Changing the variable back to $n$, this is equal to $4(n+1)^2-(n+1)$, as required.
A: Your steps so far are all correct. There is just one step to go: 
So you want to get $4(n+1)^2 - (n+1)$ and have something like $(n+1) \cdot X$ with $X$ the term you computed. Rewrite what you want to get in this form: 
$$ 4(n+1)^2 - (n+1) = (n+1) \cdot \bigl(4(n+1) - 1\bigr) $$
This gives you (writing now another way round): 
\begin{align*} 
   (n+1)(4n+3) &= (n+1)(4n+4 - 1)\\ 
    &= (n+1) \cdot \bigl(4(n+1) - 1\bigr) \\
    &= 4(n+1)^2 - (n+1)
\end{align*}
as wished.
