Question

factor 225-(x+8)^2

Answer 1

Do you know the formula of the square difference?
\[a^2-b^2=(a-b)(a+b)\]

Answer 2

no i wasnt aware of that one

Answer 3

Well you should use this formula, you can rewrite your expression as:
\[15^2-(x+8)^2\]
Now you have a=15, and b=x+8. Tell me what you think it would look like after factorization?

Answer 4

(15-x+8)(15+X+8) ??

Answer 5

You're smart. You just made a little mistake with a sign. Deal with x+8 as a combination at first. So when you subtract, a minus sign will go to both x and 8.

Answer 6

HMM not sure i understand that one, thank you for the compliment btw

Answer 7

Well, see here:
\[15^2-(x+8)^2=(15-(x+8))(15+(x+8))=(15-x-8)(15+x+8)\]
\[=(7-x)(23+x)\]
And yes you really are good!

Answer 8

ohhh i see it now witht the extra parens lol ok, haha and thanks again but your really way smarter man

Answer 9

hey i have another one to ask if can help me still?

Answer 10

Yeah sure, just one :)

Answer 11

ok haha its Cos((pi/2)-x)/sin((pi/2)-x)

Answer 12

\[\cos(\pi/2-x)\div \sin(\pi/2-x)\]

Answer 13

i know i have to use the sum and difference id's for sin and cos but im hitting a wall

Answer 14

You want to simplify:
\[{\cos({\pi \over 2}-x) \over \sin({\pi \over 2}-x)}?\]

Answer 15

yea the question says find

Answer 16

It says find? A bit strange. Anyway, it clearly wants you to apply the following two formulas:
\[\cos ({\pi \over 2}-x)=\sin x ..(1)\]
\[\sin({\pi \over 2}-x)=\cos x .. (2)\]

Answer 17

Are you following?

Answer 18

i think, would it come out to tangent?

Answer 19

Exactly!! Didn't I just I say you're smart?!

Answer 20

haha i try

Answer 21

So, applying the two formulas gives:
\[{\sin x \over \cos x}=\tan x\]
I don't know if it's giving any values for x.

Answer 22

no it was just asking for an equal trig function, i didnt know those equations though, putting them in my notes, thanks man, youve helped me alot!!

Answer 23

You're welcome. I am sure you're very good at math, and you pick things quickly. Good luck!!