How do you factor the expression #12t^8-75t^4#?

1 Answer
Oct 17, 2016

Answer:

#12t^8-75t^4 = 3t^4(2t^2-5)(2t^2+5)#

#color(white)(12t^8-75t^4) = 3t^4(sqrt(2)t-sqrt(5))(sqrt(2)t+sqrt(5))(2t^2+5)#

Explanation:

#color(white)()#
Difference of squares

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We will use this a couple of times.

#color(white)()#
Factor #bb(12t^8 - 75t^4)#

First note that both terms are divisible by #3t^4#, so separate that out as a factor first:

#12t^8-75t^4 = 3t^4(4t^4-25)#

#color(white)(12t^8-75t^4) = 3t^4((2t^2)^2-5^2)#

#color(white)(12t^8-75t^4) = 3t^4(2t^2-5)(2t^2+5)#

#color(white)(12t^8-75t^4) = 3t^4((sqrt(2)t)^2-(sqrt(5))^2)(2t^2+5)#

#color(white)(12t^8-75t^4) = 3t^4(sqrt(2)t-sqrt(5))(sqrt(2)t+sqrt(5))(2t^2+5)#

The remaining quadratic #(2t^2+5)# does not factor further with Real coefficients, since #2t^2+5 >= 5 > 0# for any Real value of #t#.