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8 January, 04:49

Ali was asked to factorise x^2y^2 + 36 - 4x^2 - 9y^2. He tried some ways of grouping terms as shown below.

x^2y^2 + 36 - 4x ^2 - 9y^2 = (x^2y^2 + 36) - (4x ^2 + 9y^2)

x^2y^2 + 36 - 4x^2 - 9y^2 = (x^2y^2 + 36 - 4x^2) - 9y^2

x^2y^2 + 36 - 4x^2 - 9y^2=x^2y^2 + (36 - 4x^2 - 9y^2)

As he could not carry out factorisation with the above groupings, he concluded that the expression could not be factorised. Do you agree with him? Why or why not?

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Answers (1)
  1. 8 January, 05:14
    0
    It can be factorised

    Step-by-step explanation:

    Given

    x²y² + 36 - 4x² - 9y² (rearranging)

    x²y² - 4x² - 9y² + 36 (factor first/second and third/fourth terms)

    = x² (y² - 4) - 9 (y² - 4) ← factor out (y² - 4) from each term

    = (y² - 4) (x² - 9)

    Both factors are a difference of squares and factor in general as

    a² - b² = (a - b) (a + b)

    Thus

    y² - 4

    = y² - 2² = (y - 2) (y + 2), and

    x² - 9

    = x² - 3² = (x - 3) (x + 3)

    Hence

    x²y² + 36 - 4x² - 9y² = (y - 2) (y + 2) (x - 3) (x + 3)
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