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right? then next?
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DONT GET IT :|
by further factorising it comes to (x+3)(x-9)=0
to find the factors of
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
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
multiply out to check: x^2 -6x-27. it works
you might try this
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
take the square root of both sides
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