Question

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?

Answer 1

It can be factorised

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)