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    MandyMarieLove's Avatar
    MandyMarieLove Posts: 70, Reputation: 7
    Junior Member
    		  
    #1

    Jul 15, 2010, 07:21 PM
    Algebra II problem solvers, with work?
    Hey, I need help with my Algebra II class. I have found math solver sites, but none give me the steps to do the problem! I don't want to be given the answers and not know what to do! Helppp?
    KISS's Avatar
    KISS Posts: 12,510, Reputation: 839
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    #2

    Jul 15, 2010, 08:09 PM

    Not quite the way to learn.

    Post away with your thoughts and possible solutions.
    MandyMarieLove's Avatar
    MandyMarieLove Posts: 70, Reputation: 7
    Junior Member
    		  
    #3

    Jul 15, 2010, 08:26 PM

    Okay, so here's my big thing.. solving quadratic equations by factoring, I know it sounds stupid.. But, I don't understand the factoring process of it. Can you help?

    Here's an example:

    Identify the solutions to the equation 2x2 + 5x = 3 when you solve by factoring.
    ISneezeFunny's Avatar
    ISneezeFunny Posts: 4,175, Reputation: 821
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    #4

    Jul 15, 2010, 08:43 PM

    solving by factoring:

    step 1: move everything to one side of the equation, leaving 0 on the other side.

    2x^2 + 5x - 3 = 0

    step 2: divide by the number in front of the highest factor (in this case 2), however, it will make this very ugly, so we'll skip that.

    step 3: find factors

    2x^2 + 5x - 3 = 0

    (2x - 1)(x + 3) = 0

    step 4: FOIL it out to see if that works

    2x^2 + 6x - x - 3 = 0

    step 5: solve

    (2x - 1)(x + 3) = 0

    2x - 1 = 0
    x + 3 = 0

    2x = 1
    x = 1/2

    x = -3
    KISS's Avatar
    KISS Posts: 12,510, Reputation: 839
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    #5

    Jul 15, 2010, 08:50 PM

    Assuming the 2x2 is 2x^2

    Start by making it equal to zero

    2x^2+5x-3 = 0

    There are only two things that can get you 2x^2 and that's a 2x and an x

    The only factors of 3 are 3 and 1; one has to be negative to end up with a -3

    try them and arrive at:

    (x+3) (2x-1) = 0 is the answer.
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