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factor completely 7x^2-12xy+5y^2
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\(7x^2-12xy+5y^2\)
yes
1st of all we w'll take lcm by multiplying 5*7
we get 35
now we will add 5 and 7 in this way that we get negative 12xy
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-7xy -5xy = -12xy
so \(7x^2-7xy-5xy+5y^2\)
now take common \(7x(x-y)~-~5y(x-y)\) now factors are \((7x-5y)(x-y)\)
@cessy18 get it?
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