How do you factor #16-a^16#?
2 Answers
(4-
Explanation:
Difference of two squares
Explanation:
The difference of squares identity can be written:
#A^2-B^2=(A-B)(A+B)#
Given:
#16-a^16#
We can start to factor this using the difference of squares identity as follows:
#16-a^16 = 4^2-(a^8)^2#
#color(white)(16-a^16) = (4-a^8)(4+a^8)#
#color(white)(16-a^16) = (2^2-(a^4)^2)(4+a^8)#
#color(white)(16-a^16) = (2-a^4)(2+a^4)(4+a^8)#
#color(white)(16-a^16) = ((sqrt(2))^2-(a^2)^2)(2+a^4)(4+a^8)#
#color(white)(16-a^16) = (sqrt(2)-a^2)(sqrt(2)+a^2)(2+a^4)(4+a^8)#
#color(white)(16-a^16) = ((root(4)(2))^2-a^2)(sqrt(2)+a^2)(2+a^4)(4+a^8)#
#color(white)(16-a^16) = (root(4)(2)-a)(root(4)(2)+a)(sqrt(2)+a^2)(2+a^4)(4+a^8)#
Sticking with real coefficients for now, we can go a little further with some of the remaining factors.
For example:
#2+a^4 = (sqrt(2)+a^2)^2-2sqrt(2)a^2#
#color(white)(2+a^4) = (sqrt(2)+a^2)^2-(2^(3/4)a)^2#
#color(white)(2+a^4) = ((sqrt(2)+a^2)-2^(3/4)a)((sqrt(2)+a^2)+2^(3/4)a)#
#color(white)(2+a^4) = (sqrt(2)-2^(3/4)a+a^2)(sqrt(2)+2^(3/4)a+a^2)#
Similarly:
#4+a^8 = (2+a^4)^2-4a^4#
#color(white)(4+a^8) = (2+a^4)^2-(2a^2)^2#
#color(white)(4+a^8) = ((2+a^4)-2a^2)((2+a^4)+2a^2)#
#color(white)(4+a^8) = (2-2a^2+a^4)(2+2a^2+a^4)#
Then:
#2-2a^2+a^4 = (sqrt(2)+a^2)^2-(2sqrt(2)+2)a^2#
#color(white)(2-2a^2+a^4) = (sqrt(2)+a^2)^2-(sqrt(2sqrt(2)+2)a)^2#
#color(white)(2-2a^2+a^4) = ((sqrt(2)+a^2)-sqrt(2sqrt(2)+2)a)((sqrt(2)+a^2)+sqrt(2sqrt(2)+2)a)#
#color(white)(2-2a^2+a^4) = (sqrt(2)-sqrt(2sqrt(2)+2)a+a^2)(sqrt(2)+sqrt(2sqrt(2)+2)a+a^2)#
Similarly:
#2+2a^2+a^4 = (sqrt(2)-sqrt(2sqrt(2)-2)a+a^2)(sqrt(2)+sqrt(2sqrt(2)-2)a+a^2)#
We can combine all of these to give a complete factorisation of
#16-a^16 = (root(4)(2)-a)(root(4)(2)+a)(sqrt(2)+a^2)(sqrt(2)-2^(3/4)a+a^2)(sqrt(2)+2^(3/4)a+a^2)(sqrt(2)-sqrt(2sqrt(2)+2)a+a^2)(sqrt(2)+sqrt(2sqrt(2)+2)a+a^2)(sqrt(2)-sqrt(2sqrt(2)-2)a+a^2)(sqrt(2)+sqrt(2sqrt(2)-2)a+a^2)#
