Find the value of $x$, for which the function $h(x) = \frac 1{(x-5)^2+4(x-5)+4}$ is undefined Picture of problem
I'm having an issue with a simple SAT question. If you can't see from the picture, I'm being asked to find the value of $x$, for which the function:
$$h(x) = \frac 1{(x-5)^2+4(x-5)+4}$$
is undefined. 
I figured that for it to be undefined, it should be equal to 1/0. This therefore means that $(x-5)^2+4(x-5)+4 = 0$. I further simplified it into 
$$(x-5)((x-5)+4) = 0$$
However, this gives me either $x = 1$ or $x = -3$, both of which are false. The answer is $3$. Why do I get $-3$, when the answer is $3$?
 A: You factorised the denominator incorrectly, leaving out the $+4$ term. The correct factorisation is  $(x-5+2)(x-5+2) = 0$, or equivalently $(x-3)(x-3) = 0$.
A: We can use $(a+b)^2=a^2+2ab+b^2$. 
By this formula we get  $$(x-5+2)^2=0$$ or
$$x-5=-2$$ or $$x=3$$
A: HINT: your denominator can be written as $$(x-5)^2+4(x-5)+4=(x-3)^2$$
A: Ahhh, I see what you are doing. In reply to user476879, you said: 

If (x-5)((x-5)+4) = -4
  then (x-5)(x-1) = -4
  in that case it's either x-5=-4 or x-1=-4
  neither of which are correct (respectively leading to x=1 or x=-3)

Nope. You can't solve for x like that. You have to get one side of the equation to equal zero first. It can't be -4, -10, 200, or anything else. It's got to be zero. Only once you get it to zero can you solve for what value of x will make it zero.
If you don't understand, please post again because this misunderstanding will cause you to have troubles with many many more SAT problems. Good Luck on it!
A: First of all, kudos to you for correctly identifying that what you need to do is find the value of $x$ that makes the denominator equal to 0. It appears to me that that this problem is designed to test the SAT taker on two components - the first being the ability to identify that $h(x)$ is undefined if and only if the denominator is equal to 0, and the second being the ability to solve algebraically for which $x$ value will make the denominator equal to 0.
All of the other correct answers certainly sufficient, but I am going to provide my method of solving as well. It is semantically no different from the other correct solutions, just different syntax, but I imagine you are in high school and when I was in high school some methods of solving equations did not click well for me and others did, so I will provide mine just in case it is the one that works for you.
$$(x-5)^2 + 4(x-5) + 4 = 0\\
\implies x^2 - 10x + 25 + 4x - 20 + 4 = 0\\
\implies x^2 - 6x + 9 = 0.$$
From here, someone with a good eye would catch that $$x^2 - 6x + 9 = (x-3)^2,$$
thus the unique solution to our problem is $x = 3$. However if a student is not as skilled in spotting factorizations, it would be just as easy to recall the quadratic equation:
$$ax^2 + bx + c = 0 \iff x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \hspace{4mm} \text{(for } a \text{ not = 0)}$$
and use this to arrive at 
$$x = \frac{6 \pm \sqrt{6^2 - 4(1)(9)}}{2} = \frac{6 \pm 0}{2} = 3.$$
This answer may seem too long for the SAT where speed matters - but that is only because I was verbose in explanation. When I initially used this method to solve your problem on paper by hand it probably took about 15 seconds. 
Hope you find the answer that works for you here on MSE and good luck on your SAT!
A: Set $y:= (x-5)$,  the denominator reads:
$D(y) = y^2 + 4y + 4$, 
Consider the function $H(y): = \dfrac{1}{D(y)}$.
$H(y)$ is not defined for $y$ values with $D(y) =0$.
Find the zeroes of $D(y)$:
$D(y) = y^2 + 4y +4 =0$.
Note: 
$D(y) = y^2 + 4y +4 = (y+2)^2 = 0$.
Hence $y = -2$ is excluded:
Domain of $H(y): \mathbb{R}$ \ {$-2$}.
Proceed to exclude the corresponding $x$ value, recall $y=x-5,$ from the domain of the original function $h(x)$. 
A: The denominator has to be zero for the chosen $x$ values of course;
$(x-5)^2+4(x-5)+4=0$
We can name $(x-5)=a$ to make things easier.
$a^2+4a+4=0$ 
That is a perfect square;
$(a+2)^2=0$
$a=-2$ so $x-5=-2 \rightarrow x=3$
