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Math Help - How do I "solve by factoring" something with an = sign?

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    How do I "solve by factoring" something with an = sign?

    How do I "solve by factoring" something with an = sign?

    Like this: 3x^2+5x=2

    ?
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    3x^2+(6x-x)-2=0\iff(3x^2+6x)-(x+2)=0, so (x+2)(3x-1)=0
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    Quote Originally Posted by deathtolife04 View Post
    How do I "solve by factoring" something with an = sign?

    Like this: 3x^2+5x=2

    ?
    3x^2 + 5x = 2

    3x^2 + 5x - 2 = 0

    This is something called the "ac method."

    Multiply the coefficient of the x^2 term by the constant term:
    3 \cdot (-2) = -6

    Now write out all the possible pairs of factors of -6:
    1, -6
    2, -3
    3, -2
    6, -1

    Now add the pairs of factors:
    1 + (-6) = -5
    2 + (-3) = -1
    3 + (-2) = 1
    6 + (-1) = 5

    If your quadratic equation factors over the integers (which is the long term for what they mean by "factoring") then the coefficient of the x term is somewhere in this list. In this case the coefficient of the x term is 5, so we have:
    6 + (-1) = 5.

    So here's the equation again:
    3x^2 + 5x - 2 = 0

    We want to rewrite the linear (x) term as 5x = 6x + (-1)x:
    3x^2 + (6x - x) - 2 = 0

    Now regroup:
    (3x^2 + 6x) + (-x - 2) = 0

    Factor each of the terms in parenthesis:
    3x(x + 2) + (-1)(x + 2) = 0

    Now each term has a common factor of x + 2, so factor that:
    (3x - 1)(x + 2) = 0

    Voila! It is factored. (You can, and should, multiply this back out to check that it matches the original expression.)

    From here we will use the property that if the product ab = 0, then either a = 0 or b = 0.

    So we know that
    (3x - 1)(x + 2) = 0

    Thus either
    3x - 1 = 0 \implies x = \frac{1}{3}
    or
    x + 2 = 0 \implies x = -2

    -Dan
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