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

1 Answer
Jan 3, 2016

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)#

Explanation:

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)#