How do you factor # x^4 - 12x^2 + 36#?

1 Answer
May 11, 2016

Answer:

#x^4-12x^2+36 = color(blue)((x+sqrt(6))^2(x-sqrt(6))^2#

Explanation:

If we temporarily replace #x^2# with #a#, we would have:
#color(white)("XXX")a^2-12a+36#
with fairly obvious factors #(a-6)^2#

Replacing the #a# with the original #x^2#
we have
#color(white)("XXX")x^4-12x^2+36=color(blue)((x^2-6)^2)#

If we regard #6# as #(sqrt(6))^2#
then #(x^2-6)# can be thought of as the difference of squares with standard factors:
#color(white)("XXX")(x^2-(sqrt(6)^2)=(x-sqrt(6))(x+sqrt(6))#

and therefore the original equation can be further factored as
#color(white)("XXX")x^4-12^x+36=[(x-sqrt(6))(x+sqrt(6))]^2#

#color(white)("XXXXXXXXXXX")=color(blue)((x-sqrt(6))^2(x+sqrt(6))^2)#