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12 July, 19:48

What is the factored form of the expression over the complex numbers?

121x2+36y2

(11x+6y) (11x-6y)

(11x+6iy) (11x-6iy)

(11x-6iy) (11x-6iy)

(11x+6iy) (11x+6iy)

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Answers (2)
  1. 12 July, 19:54
    0
    (11x + 6iy) (11x - 6iy)

    using the difference of squares identity

    a² - b² = (a - b) (a + b)

    a² + b² = (a - bi) (a + bi)

    121x² = (11x) ² ⇒ a = 11x and 36y² = (6y) ² ⇒ b = 6y

    121x² + 36y² = (11x - 6iy) (11x + 6iy)
  2. 12 July, 19:55
    0
    121x² + 36y²

    = 121x² - (-36y²) Notice that the first and last term are perfect squares.

    = (11x - 6iy) (11x + 6iy) So, you can factor using the difference of squares.

    Answer: B
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