Question

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

Answer 1

(x-3)(x-3)
x^2-3x-3x+9
x^2-6x+9=36
x^2-6x-27

right? then next?

Answer 2

Sum = -6

Product = -27

So the No are -9 and 3

Answer 3

so

x^2-6x-27 = x^2 + 3x - 9x - 27 =0 Nw Factorize

Answer 4

?

Answer 5

DONT GET IT :|

Answer 6

by further factorising it comes to (x+3)(x-9)=0

Answer 7

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

Answer 8

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

Answer 9

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)

Answer 10

if you are looking for the zeros, then there is no reason to factor, you find them in two steps, as @phi wrote above

Answer 11

thanks guys!