Question:

4y²-8y-12? I think I might almost have it down pat.?

by  |  earlier

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I have got 2(y²-4y-6) is that right? And can you factor even lower?

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  1. (4y-12) (y+1)

    4(y-3)(y+1)


  2. 4(y^2-2y-3)

    4(y-3)(y+1)

  3. 4y²-8y-12

    4(y²-2y-3)

    4(y - 3)(y + 1)

  4. No you aren't right...let me correct you:

    4y^2 - 8y - 12

    4 ( y^2 - 2y - 3 )      {taking common}

    4 ( y^2 - 3y + y -3 )     {breaking the middle term}

    4 [y (y - 3 ) + 1 ( y - 3) ]      {taking common}

    4(y - 3) (y+1) ANSWER

    But tell me one thing .... how did you wrote the power of y like that??? Help in additional information please

  5. not quite:

    4y²-8y-12

    =4 (y^2 - 2y -3)

    =4 (y-3) (y+1)

    ...

    /*********************************/

    and if you set it = 0

    ie:

    4y²-8y-12 = 0

    4(y-3) (y+1) = 0

    (y-3) (y+1) = 0

    means

    y = 3 or y= -1 are "zeros" of the equation 4y²-8y-12 = 0

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