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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?

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4 Answers 4

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 $$\begin{align} 4(n+1)^2-(n+1)&=4(n^2+2n+1)-(n+1)\\&=(4n^2+8n+4)-(n+1)\\&=4n^2+7n+3.\end{align}$$

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Hm, although it does prove that both are equal, shouldn't I begin from $4n^2+7n+3$ to $4(n+1)^2-(n+1)$ instead of viceversa as it seems in your answer? –  Zol Tun Kul Nov 26 '12 at 8:35
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@Omega: If you prove that $B=A$, you have proved that $A=B$. –  André Nicolas Nov 26 '12 at 8:37
    
Well, yeah, but I do mean I'd like to see how to go from A to B instead of B to A purely for exercise purposes. It isn't about proving the equality of their values, but the form/arrangement itself of the expressions what I'm mostly interested in. –  Zol Tun Kul Nov 26 '12 at 8:39
    
Well, we can cheat, and work sort of backwards. $4n^2+7n+3=4n^2+8n -n+3$. But $4n^2+8n=4(n^2+2n)$. Now $n^2+2n$ is almost the square of $n+1$. Put in a $1$, you get $4(n^2+2n+1)$. This means we have added $4$ to the original expression. So we need to subtract it. Our original expression is therefore equal to $4(n+1)^2-n+3-4$. this is equal to $4(n+1)^2-(n+1)$. –  André Nicolas Nov 26 '12 at 8:46
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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: $$\begin{aligned}4n^2-n+(8n+3)&=4n^2+7n+3\\&=4n^2+8n+4-n-1\\&=4(n^2+2n+1)-n-1\\&=4(n+1)^2-(n+1)\end{aligned}$$

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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.

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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.

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