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factorise x^2-4xy+4y^2
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\[x^2-4xy+4y^2=x^2-4xy+(2y)^2\] By \[a^2+2ab+b^2=(a+b)^2\], \[x^2-4xy+4y^2=x^2-4xy+(2y)^2=(x-2y)^2\]
I've never heard of "factorise."
Factor the following: \(x^2-4 x y+4 y^2\) Factor the quadratic \(x^2-4 x y+4 y^2\) The factors of 4 that sum to -4 are -2 and -2. So, \(x^2-4 x y+4 y^2 = (x-2 y) (x-2 y):\) \[(x-2 y) (x-2 y) \]Combine products of like terms. \[(x-2 y) (x-2 y) = (x-2 y)^2: \]
Which leaves you with..?
Thanks guys.
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