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Question

Question: Factorise: \((1 - {x^2})(1 - {y^2}) + 4xy\)...

Factorise: (1x2)(1y2)+4xy(1 - {x^2})(1 - {y^2}) + 4xy

Explanation

Solution

To solve questions like these that involve factorisation of polynomial functions, we have a very specific approach. We first expand the expression until we get a form that can be separated and taken common to give separate factors.

Formula used: This expression requires the formula for the subtraction of squares of terms that gives you individual factors where one is sum of terms and another is difference of sums.
a2b2=(ab)(a+b){a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)
Another formula used here is the expansion (or in this case, contraction of) the squares of the sum of two terms.
That is represented mathematically by:
(a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab

Complete step-by-step answer:
Our very first step is to expand the expression by multiplying the individual terms inside the bracket to give four separate terms.
(1x2)(1y2)+4xy(1 - {x^2})(1 - {y^2}) + 4xy
Multiplying terms inside brackets to get proper polynomials without any individual factors
(1x2y2+x2y2)+4xy\Rightarrow (1 - {x^2} - {y^2} + {x^2}{y^2}) + 4xy
Now we separate the terms and arrange it in ways that can give us proper whole squares so as to get factors of this algebraic expression. We do this by taking the terms that have xx and yy as products together to make a whole square involving xyxy. For the other whole square, we make a whole xx and yy separately.
This can be illustrated mathematically as,
(12+x2y2+2xy)(x2+y2xy)\Rightarrow ({1^2} + {x^2}{y^2} + 2xy) - \left( {{x^2} + y - 2xy} \right)
This gives rise to perfect squares now that can be represented by
(1+xy)2(xy)2\Rightarrow {(1 + xy)^2} - {\left( {x - y} \right)^2}
Using the first formula mentioned above,
(1+xy+xy)(1+xyx+y)\Rightarrow (1 + xy + x - y)\left( {1 + xy - x + y} \right)

Hence, the expression factorises to this.

Note:
Expansion of polynomials is simply a matter of multiplying and separating until it is a simplified term. Then we have to decide for ourselves what terms need to be grouped to give the proper squares.