I just got to college, and right now Calculus is looking like it's going to kill me, despite math being my strong subject. Got my first test back and I think it was roughly a 72%, mainly because I was missing a lot of Trig problems. I think a lot of that is due to things that we didn't have to memorize in high school that now need to be memorized, but even still I know I need work/help with it. So I'm just going to use this thread to run off random questions from my math.

Not sure why I couldn't have just simply asked my questions without all that, but whatever, here we go:

Here's the expression:

(x-2)(x+5)

----------- ≤ 0

(x+3)

My professor has been stressing that we have two major ways to approach these problems, algebraically and geometrically. So far I've been focusing in on algebraically, because I understood that method and was lost on the geometric, but for this problem the number of cases involved makes geometric the way to go, so she said. I copied down her work to go over (I was working through a previous problem as she worked this) but am still lost on part of it. Here's how she did it:

[original problem]

x-2 = 0

x = 2

OR

x + 5 = 0

x = -5

OR

x+3 = 0

x=-3

Up to this point, I follow her. She's defining what the zeroes are. I'm not entirely sure why though. Any explanation there would be nice.

Then, she does this (she also mentioned some chart-method, but she doesn't use that so idk what she means or if that would be easier for me):

There's a number line with -5, -3, and 2 marked. -5 and 2 are closed circles, -3 is open. (ignore the periods)

<---(-5)--(-3)--(2)--->

....N| N |... N |...P

....N| P |... P |...P

....N| N |... P |...P

------------------------

....N| P |... N |...P

that has something to do with what is positive and what is negative, and from there she got the solution:

sol: (-infinity, -5] union (-3, 2]

which I understand to be "x is less than or equal to -5 or x is greater than -3 but less than 2"

My main problem is understand what the little number line/letters thing is and how I determine which is negative and which is positive.

Ok, now the next problem, we started with:

xÂ² > 4

now the work:

xÂ² > 4

xÂ² - 4 > 0 (why did we do that?)

(x-2)(x+2) > 0

case 1:(x-2)(x+2) > 0

x-2 > 0 AND x+2 > 0

x>2 AND x>-2

THEREFORE x>2

case 1 I understand just fine, it's case 2 where she lost me:

case 2:(x-2)(x+2) > 0

x - 2 < 0 AND x+2 < 0 (and here I'm lost, how did we get from the first line of this case to this?)

x<2 and x<-2

sol: x{x|x<-2 OR x>2}