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I have this expression:

$$4n^2-n+(8(n+1)-5)$$

And I know it is equivalent to this:

$$4(n+1)^2-(n+1)$$

I need to simplify my expression to get the same structure as that one. However, no matter what I try, I don't end up with that structure.

My question is, well, how to reach that. But truly, the real problem is that I seem to lack this sense of "knowing which method to use to simplify to get the structure I want". Is there some kind of rule or guide about this? Or is it all about practice and experience?

Edit

If anyone was curious, I ended up with this: $$4n^2+7n+3$$ Yes, I did multiplications and additions. Couldn't find common factor (as far I can tell) or any other trick.

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A useful technique when your answer is a complicated expression is working backwards. Try expanding $4(n+1)^2−(n+1)$ and see if you end up with $4n^2+7n+3$. If you do, see if you can follow the steps back up the chain to finish the problem. –  Eugene Shvarts Jul 5 '12 at 2:59
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5 Answers

up vote 0 down vote accepted

To factor $\rm\:f = 4n^2+7n+3\:$ note $\rm\,a\,x^2 + (a\!+\!c)\,x + c = ax\,(x+1) + c\,(x+1) = (ax+c)\,(x+1).\,$ Alternatively, as in your prior question, apply the AC-method as follows:

$$\begin{eqnarray}\rm 4f\, &=&\rm\ \ 16n^2 +\ 7\cdot 4n\, +\, 4\cdot 3 \\ &=&\rm\ (4n)^2 + 7\,(4n)\, + 4\cdot 3 \\ &=&\rm\ \ N^2\ +\ 7\, N\ +\ 12\quad for\quad N = 4n \\ &=&\rm\ (N\ +\ 4)\,(N\ +\ 3) \\ &=&\rm\ (4n\, +\, 4)\,(4n\, +\, 3) \\ \rm f\, &=&\rm\ (\ n\ +\ 1)\,(4n\, +\, 3) \end{eqnarray}$$

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With the AC-method, I'm still not sure how would I reach $$4(n+1)^2-(n+1)$$ –  Omega Jul 5 '12 at 4:13
    
@Omega $ $ Hint $\rm\ 4n+3 = 4(n\!+\!1 - 1)+3 = 4(n\!+\!1) -1.\,$ Now multiply that by $\rm\,n\!+\!1.\qquad$ –  Bill Dubuque Jul 5 '12 at 4:21
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Maybe we can factor your expression $4n^2+7n+3$. Fairly quickly we get $(n+1)(4n+3)$. That seems nice enough already. But one can observe that $4n+3=4(n+1)-1$. Then your expression becomes $(n+1)[4(n+1)-1$, which can also be written as $4(n+1)^2-(n+1)$.

One reason that we know that $4n^2+7n+3$ will factor nicely is that the polynomial $4x^2+7x+3$ has the root $x=-1$. Thus $x+1$ must divide the polynomial. Do the division, using the ordinary division process for polynomials, which is much like the division process for integers. The quotient turns out to be $4x+3$.

Remark: It is awkward to say anything in general. After working with many particular examples, one accumulates a set of tools that often turn out to be useful in new settings.

There is a general procdure for expressing a polynomial $P(x)$ as a polynomial $Q(x-a)$, where $a$ is any given number. The easiest description involves the calculus, but one can also give a purely algebraic description of the process.

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Since first term is $4(n+1)^2$, try to extract this term from $4n^2+7n+3$ by addition and subtraction of terms, e.g $4n^2+7n+3=4n^2+8n+4 -n-1$, this way you got first three terms of the expression giving $4(n+1)^2$ and the term left is $-(n+1)$ which is of course the second term required, therefore, $4n^2+7n+3=4(n+1)^2-(n+1)$.

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Try doing it directly, like this:

$$4n^2-n+(8(n+1)-5) = 4n^2+8(n+1)-n-5=4n^2-4+8(n+1)-n-1$$

Then note that $4n^2-4=4(n^2-1)=4(n+1)(n-1)$ so that the expression becomes:

$$4(n+1)(n-1)+8(n+1)-(n+1) = 4(n+1)(n-1+2)-(n+1) = 4(n+1)^2-(n+1)$$

where I have extracted the common factor $4(n+1)$ from the first two terms.

The strategy for doing this was to isolate the $(n+1)$ term at the end of the given target expression and to work with the rest.

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I don't see where this needs a trick. There is nothing wrong with simplifying each expression and calling them equal if, and only if, they simplify the same. Here is the un-tricky computation.

$$\begin{align} 4 n^2 - n + ((8(n + 1) - 5) &= 4 n^2 - n + ((8 n + 8) - 5) \\ &=4 n^2 - n + (8 n - 3) \\ &=4 n^2 + 7 n + 3 \end{align}$$

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

If you need to make this sound like it's a fancy trick, call it "linearity of polynomials in real numbers" (or complex numbers, for that matter). That phrase refers to the principle that we used, that if $p$ is a real-valued polynomial function of one real variable and $a$, $b$, and $x$ are real numbers, then $p(a\cdot x + b) = a\cdot p(x) + b$.

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