5
$\begingroup$

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?

$\endgroup$
6
  • $\begingroup$ It is the canonical expression $\endgroup$
    – Maman
    Aug 9, 2014 at 15:09
  • 25
    $\begingroup$ 4 identical answers to this question is just not enough, can someone add another one?! $\endgroup$
    – Winther
    Aug 9, 2014 at 15:13
  • 5
    $\begingroup$ Use $(a-b)^2=a^2-2ab+b^2$. This is easy middle school math. $\endgroup$
    – user5402
    Aug 9, 2014 at 15:55
  • 2
    $\begingroup$ 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. $\endgroup$
    – Eff
    Aug 9, 2014 at 15:57
  • 4
    $\begingroup$ @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. $\endgroup$ Aug 9, 2014 at 15:57

5 Answers 5

32
$\begingroup$

Same as others but with some colors

$(x-2)^2=\color{red}{(x-2)}\color{blue}{(x-2)}$

$\color{red}{(x-2)}\color{blue}{(x-2)}=\color{red}x\color{blue}{(x-2)}\color{red}{-2}\color{blue}{(x-2)}$

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

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

$\endgroup$
1
  • 1
    $\begingroup$ Thank you all upvoters :D. @JohnRennie, thank you for the comment ;) $\endgroup$
    – Vikram
    Aug 10, 2014 at 5:38
10
$\begingroup$

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

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 $x = 2\pm 2\sqrt 3$.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
3
$\begingroup$

$x^{2}-4x-8$=$(x-2)^{2}-4-8$=$(x-2)^{2}-12$

$\endgroup$
0
$\begingroup$

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

$$(x-2)^2-12=x^2-4x+4-12=x^2-4x-8$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .