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stupidinmath
how to factor .. (x-3)^2=36
(x-3)(x-3) x^2-3x-3x+9 x^2-6x+9=36 x^2-6x-27 right? then next?
Sum = -6 Product = -27 So the No are -9 and 3
so x^2-6x-27 = x^2 + 3x - 9x - 27 =0 Nw Factorize
by further factorising it comes to (x+3)(x-9)=0
to find the factors of x^2-6x-27 first notice that the 27 is negative. that means you are looking for 2 numbers with different signs (either + and - or - and +) List the factors of 27 1,27 3,9 do any of the pairs give -6 , when the numbers have different signs? 1 and 27 could be -1 + 27 or +1-27 but neithert give -6 but +3-9 = -6 so we want +3 and -9 (x+3)(x-9) multiply out to check: x^2 -6x-27. it works
you might try this \[(x-3)^2=36\] \[(x-3)^2-36=0\] then since \(a^2-b^2=(a+b)(a-b)\) you get \[(x-3+6)(x-3-6)\] and then add the numbers
and not to confuse you, you could have started with (x-3)^2=36 take the square root of both sides x-3= ±6 and x= +6+3= 9 and x= -6+3= -3 this means the factors are (x-9)(x+3)
if you are looking for the zeros, then there is no reason to factor, you find them in two steps, as @phi wrote above