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Math Help - solving this

  1. #1
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    solving this

    its pre straight forward, ive got an idea on how to do it, but i keep getting the wrong answer.

    -6x^2+13x-5

    so far i got (3x- ?)(2x+?)
    i know how to solve it, but im getting the wrong numbers in the question marks above. ive got the idea of getting rid of the 6x^2 by x it by constant 5.
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  2. #2
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    Quote Originally Posted by johnsy123 View Post
    its pre straight forward, ive got an idea on how to do it, but i keep getting the wrong answer.

    -6x^2+13x-5

    so far i got (3x- ?)(2x+?)
    i know how to solve it, but im getting the wrong numbers in the question marks above. ive got the idea of getting rid of the 6x^2 by x it by constant 5.
    1. You can't solve a term. I assume that you want to factor the given term ... ? If so:

    -6x^2+13x-5 = -\left(6x^2-13x+5  \right)

    Since 13 = 2 \cdot 5 + 3 you can factor this term into:

    -\left(6x^2-13x+5  \right) = -(2x-1)(3x-5)
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  3. #3
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    Quote Originally Posted by johnsy123 View Post
    its pre straight forward, ive got an idea on how to do it, but i keep getting the wrong answer.

    -6x^2+13x-5

    so far i got (3x- ?)(2x+?)
    i know how to solve it, but im getting the wrong numbers in the question marks above. ive got the idea of getting rid of the 6x^2 by x it by constant 5.
    Assuming the expression is = 0...

    -6x^2 + 13x - 5 = 0

    -(6x^2 - 13x + 5) = 0


    You need to find two numbers that multiply to become  6 \cdot 5 = 30 that also add to be -13.

    I think you'll find that the numbers are -3 and -10.

    So break up the middle term then factor by grouping:


    -(6x^2 - 3x - 10x + 5) = 0

    -[3x(2x - 1) - 5(2x - 1)] = 0

    -(2x - 1)(3x - 5) = 0.


    Now use the Null Factor Law:

    2x - 1 = 0 or 3x - 5 = 0.

    So x = \frac{1}{2} or x = \frac{5}{3}.
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  4. #4
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    Quote Originally Posted by johnsy123 View Post
    its pre straight forward, ive got an idea on how to do it, but i keep getting the wrong answer.

    -6x^2+13x-5

    so far i got (3x- ?)(2x+?)
    i know how to solve it, but im getting the wrong numbers in the question marks above. ive got the idea of getting rid of the 6x^2 by x it by constant 5.
    You can always use the quadratic formula, where

    a = -6

    b = 13

    c = -5
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