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9 July, 09:36

3r (2p-5) - p (2p-5) factorise it

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  1. 9 July, 09:42
    0
    3r (2p - 5) - p (2p - 5)

    The common factor is (2p - 5), therefore we can write:

    (2p - 5) (3r - p)
  2. 9 July, 12:47
    0
    3r (2p - 5) - p (2p - 5)

    let a = 2p - 5

    3r (2p - 5) - p (2p - 5) = 3ra - pa = a (3r - p)

    let's substitute back the value of a.

    a (3r - p) = (2p - 5) (3r - p)

    Therefore: 3r (2p - 5) - p (2p - 5) = (2p - 5) (3r - p)
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