Help with algebra review for geometry class?

Question

okay, there's a few questions i forget from algebra last year, who ever has the shortest but most informative answers gets 10 points!!!plz help me!!!

1. In general, how do you solve a quadratic equation in factored form like this: (x+3)(2x-5)=0

(don't solve equation!!!explain how)

2. The following equation is a quadratic equation that cannot be factored: x^2+3x-6=0

how do you know if an equation can be factored or not?

Answer 1

put each set in parenthesis set to zero

x+3 = 0     and  2x-5 = 0

solve for x

no to factors = -6  and add up to 3

Answer 2

1.  Set both sides = to 0 and solve for x

so x=-3 or x = 5/2

2.  You pretty much just have to look at it.  Since 6 is neg and 3 is pos, you're looking for 2 numbers that multiply to 6 and add to 3.  The factors of 6 are 1,6 or 2,3.  Neither of those can add or subtract to 3.

Answer 3

Alright, so you would do the following:

1. Use FOIL(first outer inner last) : x*2x, then x*-5, 3*2x, then 3*-5. The answer would be 2x^2 +x-15 = 0

2. What 2 numbers will multiply to become -6 and add to become 3? That is the key I always found that helped me in these problems. So, x^2+3x-6=0 cannot be factored because there are NO to numbers that will do this. Now in (x-3)^2, 3*3=9, and 3+3=6