How do you factor ${(x + y)}^{2} - 4 x - 4 y + 4$ $(x+y)^2-4x-4y+4$ ?

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How do you factor ${(x + y)}^{2} - 4 x - 4 y + 4$ $(x+y)^2-4x-4y+4$ ?

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Answer 1 · 2016-08-24T14:00:48

${(x + y - 2)}^{2}$ $(x +y-2)^2$

Explanation:

${(x + y)}^{2} - 4 x - 4 y + 4$ $(x +y)^2-4x-4y+4$
Well you have to do something...multiplying out the brackets doesn't help...
Look again... $- 4 x - 4 y = - 4 (x + y)$ $-4x-4y =-4(x +y)$
So I can write the expression as
${(x + y)}^{2} - 4 (x + y) + 4$ $(x +y)^2-4(x +y)+4$
Let's write $z = (x + y)$ $z=(x +y)$
So now we are factorising $z^{2} - 4 z + 4$ $z^2-4z+4$
$(z - 2) (z - 2)$ $(z-2)(z-2)$
Or ${(x + y - 2)}^{2}$ $(x +y-2)^2$

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How do you factor ${(x + y)}^{2} - 4 x - 4 y + 4$ $(x+y)^2-4x-4y+4$ ?

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How do you factor ${(x + y)}^{2} - 4 x - 4 y + 4$ $(x+y)^2-4x-4y+4$ ?