How do you factor # 1 - 2.25x^8#?

Question

Abigail Allen · Accepted Answer

#1-2.25x^8#
#=1/4(root(4)(2)-root(4)(3)x)(root(4)(2)+root(4)(3)x)(sqrt(2)+sqrt(3)x^2)(sqrt(2)-root(4)(24)x+sqrt(3)x^2)(sqrt(2)+root(4)(24)x+sqrt(3)x^2)#

Some identities that we will use: Difference of squares identity: #a^2-b^2 = (a-b)(a+b)# Sum of fourth powers identity: #a^4+b^4 = (a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2)# So: #1-2.25x^8# #=1/4(4-9x^8)# #=1/4(2^2-(3x^4)^2)# #=1/4(2-3x^4)(2+3x^4)# #=1/4((sqrt(2))^2-(sqrt(3)x^2)^2)(2+3x^4)# #=1/4(sqrt(2)-sqrt(3)x^2)(sqrt(2)+sqrt(3)x^2)(2+3x^4)# #=1/4((root(4)(2))^2-(root(4)(3)x)^2)(sqrt(2)+sqrt(3)x^2)(2+3x^4)# #=1/4(root(4)(2)-root(4)(3)x)(root(4)(2)+root(4)(3)x)(sqrt(2)+sqrt(3)x^2)(2+3x^4)# #=1/4(root(4)(2)-root(4)(3)x)(root(4)(2)+root(4)(3)x)(sqrt(2)+sqrt(3)x^2)((root(4)(2))^4+(root(4)(3)x)^4)# #=1/4(root(4)(2)-root(4)(3)x)(root(4)(2)+root(4)(3)x)(sqrt(2)+sqrt(3)x^2)(sqrt(2)-sqrt(2)root(4)(2)root(4)(3)x+sqrt(3)x^2)(sqrt(2)+sqrt(2)root(4)(2)root(4)(3)x+sqrt(3)x^2)# #=1/4(root(4)(2)-root(4)(3)x)(root(4)(2)+root(4)(3)x)(sqrt(2)+sqrt(3)x^2)(sqrt(2)-root(4)(24)x+sqrt(3)x^2)(sqrt(2)+root(4)(24)x+sqrt(3)x^2)#

Kennedy Adams · Answer

undefined To factor the expression $1 - 2.25x^8$, you can rewrite it as $1 - (1.5x^4)^2$, which is the difference of squares. So, factoring it yields $(1 - 1.5x^4)(1 + 1.5x^4)$.