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"In the complete factorization of $y^2 - 4 - x^2 + 4x$, one of it's factors is"

A) $x - 4$

B) $y + 2$

C) $y - x - 2$

D) $y - x + 2$

I did a little calculator trick I always use. I got it wrong. I've decided to study more deeply the steps required to properly answer this manually.

However, unfortunately I couldn't understand what someone explained to me a while back. Can someone enlighten me here? I find it puzzling how to get the complete factorization of such thing.

Btw, I chose D. Got it wrong. I believe it was C after some extra calculation, but I am not sure anymore...

Thanks.

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1  
Hints: can you factorize $x^2-4x+4$? And $y^2-z^2$ for any $z$? –  Did May 9 '11 at 5:33
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Hint: You are looking at $y^2-(x-2)^2$. Now recall the factorization of $u^2-v^2$. –  André Nicolas May 9 '11 at 5:33

3 Answers 3

up vote 4 down vote accepted

D is the correct answer:

\begin{align*} y^2-4-x^2+4x &= y^2- (x^2-4x+4)\\ &=y^2-(x-2)^2\\ &=\left(y+(x-2) \right) \left(y-(x-2) \right)\\ &=(y+x-2)(y-x+2) \end{align*}

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I had made corrections using align environment so that it looks neat. –  user9413 May 9 '11 at 5:35
    
Please edit it. –  user9413 May 9 '11 at 5:35
    
Whoa, I wouldn't have been able to figure those steps you did. Is there are... specific order of... instructions I need? –  Zol Tun Kul May 9 '11 at 5:53
    
@Omega: The key idea for this problem occurs on the first line - it is to group the terms with the same variable (keep all the $y$'s on one side, and move all the $x$'s to the other, adding in negatives where necessary so we can group all the $x$'s in one big expression surrounded by parentheses. On the second line, we notice that the term consisting of only $x$'s can be factored itself; there is no trick here, you just have to practice noticing that $x^2-2ax+a^2=(x-a)^2$ for any number $a$. After that, you can use another rule you should be familiar with, the difference of squares rule. –  Zev Chonoles May 9 '11 at 6:00
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This is all hinted at in the comments posted underneath your question. Now, the process we used for this problem is by no means universal; but what I would advise in general, so that you are able to come up with this kind of answer yourself in the future, is to remember the important factoring formulas (such as $a^2-b^2=(a+b)(a-b)$), especially understanding why they work, and (this next bit is key) being comfortable noticing a factorization pattern like $a^2-b^2$ or $(a+b)(a-b)$ when $a$ and $b$ might be complicated expressions themselves. –  Zev Chonoles May 9 '11 at 6:03

Complete the square in $x$ and everything should be clear.

Not that I particularly encourage this, but you can rule out each of the choices by picking values of $x$ and $y$ such that exactly one of the four choices is equal to zero and then plugging those values into the original expression. For the four choices you can pick $(x, y) = (4, 0), (1, -2), (0, 2), (2, 0)$ respectively.

Alternately, since all of the factors are linear you can just solve for one variable in one of the linear factors and then plug the corresponding expression into the original expression to see if it equals zero.

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HINT $\ $ If $\rm\ deg_{\:y}\ c = 0\ $ then $\rm\:\ y^2 - c\ $ is reducible $\rm\: \iff\ c = b^2$

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