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Math

posted by on .

I'm totally stuck on this problem!!

Simplify:
(2a^2+13a-7/3a^3-27a)(4a^2-1/9a^2)

  • Math - ,

    This probably should have been written as
    (2a^2+13a-7)/[(3a^3-27a)(4a^2-1)/9a^2]


    Try factoring 2a^2 +13a -7 into
    (2a -1)(a+7)
    and 3a^3-27a into
    3a(a^2-9 = 3a(a+3)(a-3)
    and 4a^2 -1 into (2a +1) (2a -1)

    The 3a and (2a-1) terms will cancel.

  • Math - ,

    No,
    I'm sure the problem is written like this:
    [(3a^3-27a)/(2a^2+13a-7)]/[9a^2/4a^2-1)]

    And to divide you multiply by the reciprocal.

    So i still don't get the problem I also know to factor, but I'm stil confused.

  • Math - ,

    What you just typed as sure as the problem as written is NOT close to what you posted. Factor, then divide out common factors.

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