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I'm going over my quadratic equations for the ACT and I came across this quadratic:

$$(x – 2)^2 – 12$$

My teacher said we could have factored it out into this:

$$x^2 – 4x – 8$$

But I just don't understand where he got the $-4x$! Help?

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It is the canonical expression –  Maman Aug 9 '14 at 15:09
4 identical answers to this question is just not enough, can someone add another one?! –  Winther Aug 9 '14 at 15:13
Use $(a-b)^2=a^2-2ab+b^2$. This is easy middle school math. –  whatever Aug 9 '14 at 15:55
Also do not fall into the trap of thinking that $(x-2)^2 = x^2-2^2$. @metacompactness Is that really necessary? No reason to mock him. –  Eff Aug 9 '14 at 15:57
@Winther It could be worse. How about a different answer that proceeds by evaluating the given polynomial at three points and then, since it's quadratic, computes it by a system of three linear equations for the coefficients. Or, worse yet, applies the Lagrange interpolation formula. –  Andreas Blass Aug 9 '14 at 15:57

5 Answers 5

up vote 11 down vote accepted

$$(x-2)^2 = (x - 2)(x-2) = x^2 - 2x -2x + (-2)(-2) = x^2 - 4x + 4$$

This is called expanding $(x-2)^2$. We factor $x^2 - 4x + 4$ when we write it as the product of its factors, in this case $(x-2)(x-2) = (x-2)^2$.

Now, $$(x-2)^2-12=(x^2-4x+4)-12=x^2-4x-8$$ You can find the zeros of the quadratic by setting $x^2 - 4x - 8 = 0 $, then using the quadratic formula, which will yield $x = 2\pm 2\sqrt 3$.

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Same as others but with some colors



$\hspace{65 pt}=\underbrace{\color{red}x\times \color{blue}x}\hspace{5 pt}+\underbrace{\color{red}x \times \color{blue}{-2}}\hspace{5 pt}\underbrace{\color{red}{-2} \times \color{blue}x}\hspace{5 pt}\underbrace{ \color{red}{-2} \times \color{blue}{-2}}$ $\hspace{70 pt}=x^2 \hspace{25 pt}-2x\hspace{15 pt}-2x\hspace{15 pt}+4$

$\hspace{65 pt}=x^2-4x+4$

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+1 for the colours :-) –  John Rennie Aug 9 '14 at 19:35
Thank you all upvoters :D. @JohnRennie, thank you for the comment ;) –  Vikram Aug 10 '14 at 5:38


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