How to factor this polynomial by grouping? -- 2xy - 9cx - 18cy + 81c2^?

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  • 1 month ago

    2xy - 9cx - 18cy + 81c^2

    group the first two terms and factor out an x:

    2xy - 9cx = x(2y - 9c)

    group the last two terms and factor out (-9c):

    -18cy + 81c^2 = -9c(2y - 9c)

    Now you have:

    x(2y - 9c) - 9c(2y - 9c)

    notice that the binomials inside the ( )'s are the same, so take the whole binomial out as a single common factor:

    (2y - 9c) (x - 9c)

    hope it helps!

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  • sepia
    Lv 7
    1 month ago

    2xy - 9cx - 18cy + 81c^2

    = (9 c - x) (9 c - 2 y)

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  • 1 month ago

     2xy - 9cx - 18cy + 81c^2

    = (9 c - x) (9 c - 2 y)

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  • I don't get what -- is supposed to mean.

    You have 3 variables:  x , y , c.  If we group the first 2 terms and the last 2 terms, we get:

    x * (2y - 9c) - c * (18y - 81c) =>

    x * (2y - 9c) - 9c * (2y - 9c) =>

    (x - 9c) * (2y - 9c)

    If we group the 1st and 3rd term and the 2nd/4th term, we get:

    2xy - 18cy - 9cx + 81c^2 =>

    2y * (x - 9c) - 9c * (x - 9c) =>

    (2y - 9c) * (x - 9c)

    Which one do you like better?

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