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4 May, 17:15

How would you solve (3a-4) ^2?

According to my teacher the answer is (3 (a-1)) ^2 = 9a^2 - 18 a + 9

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Answers (2)
  1. 4 May, 17:21
    0
    Well (3a - 4) ^2 is equivalent to (3a - 4) * (3a - 4)

    How I would solve that is by firstly multiplying (3a - 4) by 3a (first part of second bracket) then adding that to (3a - 4) * - 4 (second part of second bracket)

    That would equal

    (9a^2 - 12a) + (-12a + 16)

    = 9a^2 - 12a - 12a + 16

    = 9a^2 - 24a + 16

    Sorry if my answer is rubbish
  2. 4 May, 17:24
    0
    (3a - 4) ^2 =

    (3a - 4) (3a - 4) =

    3a (3a - 4) - 4 (3a - 4) =

    9a^2 - 12a - 12a + 16 =

    9a^2 - 24a + 16 < = = I am not getting what ur teacher got

    is the problem written correctly?

    Because (3a - 3) ^2 = 9a^2 - 18a + 9
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