Please can you help me with factors?

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how do solve:

x^2+2x-15

I have tried everything and if you know please I'm literally crying my eyes out over this! I am that desperate I have a major exam in 2 days!

x^2+2x-15=0

Add 15 to both sides
x^2+2x-15+15=0+15

simplify
X^2+2x=15

Factor left side
x(x+2)=15

Factor right side (what times what = 15)
One of the factors = X

x^2 + 2x - 15 = 0
choose a, b such that a*b = -15 and a+b = 2;
=> (x+a)(x+b) = 0
x = -a or x = -b.

Since there are only a handful of factorizations for -15, you can solve this by trial and error.

Oops, I made a detailed post here, but I was wrong. The people above me have the right idea.

(x - 3)(x + 5)

Since x(x) = x^2, 5x - 3x = 2x, and -3(5) = -15.

Okay. First look at -15 and look at that x^2. That x^2 is the same as (1)x^2. So, 1 * -15 is -15. We need two numbers that when multiplied, it gives you -15 and when added, it gives you 2. Hmm. All of the factors of 15 are: -1,15 or 1, -15 (since -1 x 15 =-15 and 1 x -15 = -15), and -3, 5 and 3,-5 (since -3 x 5 = 15 and 3 x -5 = -15). (Whew, huh?)

Let's see now - we need these two factors to add up to 2. Well, -1 + 15 = 14, and 1 + -15 = -14. Darn! (You seeing spots yet? Go to Doctor I.C. Spots) Sorry for that commercial. Now 3 + -5 is -2, and -3 + 5 = 2 = hey!! ! That's it!! ! So, our factors are -3 and 5, because when you add those two together, the sum is 2. Now...

(x-3)(x+5) = 0 (you plug in the factors including the signs into the parenthesis)

This means either x-3 = 0 or x+5 = 0; solving both gives you x = -3, and x = 5. Ta da! (C# major chord). I hope that helps you.

To review: find out all factors, put them in pairs (1,-15) (-1,15), (-3, 5), and (3,-5), and figure out what times what AND what and what adds up to that middle term. Then put those correct factors in parenthesis (x + )(x + ), and solve each one for 0.

Have a nice day.

Thanks! man you saved my life! That's exactly what I got! Thanks for your help everyone!

for those, the easiest method is probably the quadratic formula. anytime you have anything of the form: a*x^2 + b*x + c, the two factors are always:

(x - q)*(x-r)

where

q = (-b + sqrt(b^2 -4*a*c))/2a
and
r = (-b - sqrt(b^2 -4*a*c))/2a

that way, if you have undetermined constants like "p" in your formula, you can stick them right in the above formulas and get a correct answer.