Is it possible to factor #y=x^2-6x+9 #? If so, what are the factors?

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Answer 1 · 2016-01-19T21:32:28

#(x-3)^2#

Imagine you haven't realized that expression is of the type:

#x^2 +2ax+ a^2#, which gives the answer automatically:
#(x+a)^2#, since it is a notable case.

In these cases you have to use the general method for solving

second degree polynomials:

#alpha=(-b +-sqrt(b^2-4ac))/(2a)#

so, it will be:

#alpha=(6 +-sqrt(6^2-4*1*9))/(2*1)#

#=(6 +-sqrt(36-36))/(2)#

#=(6+-0)/2#

which can be devided in

#alpha1=(6+0)/2# and #alpha2=(6-0)/2#

#alpha1=3# or #alpha2=3#

So the factorization is

#(x-alpha1)(x-alpha2)#

#(x-3)(x-3)# or #(x-3)^2#