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    phoenirius's Avatar
    phoenirius Posts: 34, Reputation: 1
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    #1

    Jan 5, 2006, 07:15 PM
    factoring for x-intercepts
    How would you find the x-intercepts of y=3x^2-6X by factoring?
    Thanks
    CaptainForest's Avatar
    CaptainForest Posts: 3,645, Reputation: 393
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    #2

    Jan 5, 2006, 07:55 PM
    y=3x^2-6X

    So, to find X-intercepts, y must equal 0

    So,

    y=3x^2-6X
    0=3x^2-6X
    0=3x(x-2)

    3x=0
    x = 0

    AND

    x-2=0
    x=2

    Therefore, x-intercepts are 0 and 2.

    Now to check. Sub the values of x into equation, y must equal 0

    y=3x^2-6X
    y=3(0)^2-6(0)
    y = 0-0
    y =0

    AND

    y=3x^2-6X
    y=3(2)^2-6(2)
    y = 12 - 12
    y = 0

    Therefore, the x-intercepts are 0 and 2
    phoenirius's Avatar
    phoenirius Posts: 34, Reputation: 1
    Junior Member
     
    #3

    Jan 6, 2006, 09:07 AM
    Thank you. So out of curiosity would 3x^2-4=0 find the same intercepts? Like would you factor it out to 3x(x-2) or 3x(x+2)?
    CaptainForest's Avatar
    CaptainForest Posts: 3,645, Reputation: 393
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    #4

    Jan 6, 2006, 03:43 PM
    Quote Originally Posted by phoenirius
    Thank you. So out of curiosity would 3x^2-4=0 find the same intercepts? Like would you factor it out to 3x(x-2) or 3x(x+2)?
    Your welcome.

    3x^2-4=0 would NOT give you the same X-intercepts.

    3x(x-2) or 3x(x+2) is NOT the factor of 3x^2-4=0

    Rather,
    3x(x-2) = 3x^2 – 6x
    3x(x+2) = 3x^2 + 6x

    To solve 3x^2-4=0 you need to use the quadratic equation.

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