how to factor .. (x-3)^2=36

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how to factor .. (x-3)^2=36

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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(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

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DONT GET IT :|
by further factorising it comes to (x+3)(x-9)=0
  • phi
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
  • phi
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
thanks guys!

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