SOLUTION: I cannot remember how to factor out problems. Please help with the following examples. I keep thinking I need to isolate the x, but I'm not sure how to begin x^2-9= 2x^2+7x-

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: I cannot remember how to factor out problems. Please help with the following examples. I keep thinking I need to isolate the x, but I'm not sure how to begin x^2-9= 2x^2+7x-      Log On


   

Question 99695: I cannot remember how to factor out problems. Please help with the following examples. I keep thinking I need to isolate the x, but I'm not sure how to begin
x^2-9=
2x^2+7x-4=
3x^3-6x^2-9x=
Found 2 solutions by mdalis, Adam:
Answer by mdalis(5) About Me  (Show Source):
You can put this solution on YOUR website!
(x+3)(x-3)
(2x-1)(x+4)
3x(x+3)(x-1)

Answer by Adam(64) About Me  (Show Source):
You can put this solution on YOUR website!
x^2-9=0 (I suppose it equals zero, but whenever you submit a problem you should do it altogether.(It might seem futile to tell anyone you have got completly nothing,but it is not :-))
Anyway you can use x^2-y^2=(x-y)*(x+y) identity. And get (x-3)*(x+3) =0 Now this expression equals zero whenever at least one of the terms in it equals zero. z-3=0 /+3
z = 3
z+3=0 /-3
z = -3
2x^2+7x-4=0 for this one you will have to use quadratic formula:x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ substituting coefficients:x+=+%28-7+%2B-+sqrt%28+7%5E2-4%2A2%2A%28-4%29+%29%29%2F%282%2A2%29+ and get results 1/2 and -4

3x^3-6x^2-9x = 0 You can factor out x and get x*(3*x^2-6*x-9)=0 and again at least one of the terms in multiplication has to equal 0. First soulution: x=0.
Second term can be factored (x-3)*(x+1) (verification is up to you) that means two roots 3 and -1. Smmarized : x= 3,(-1) and 0