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How do you factor completely #x^2 - 6x + 8 #?

1 Answer

Apr 29, 2016

For #x^2-6x+8=0#
#0=x^2-4x-2x+8#
#0=x(x-4)-2(x-4)#
#0=(x-4)(x-2)#

#(x-4)=0# OR #(x-2)=0#
#x=4# OR #x=2#

For #x=4 and 2#
#x^2-6x+8# is zero
Hence #4 and 2# are its factors

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