Solve the following equation #x^8-10x^4+9 = 0#?
1 Answer
Explanation:
Given:
#x^8-10x^4+9=0#
Note that this is effectively a quadratic in
#(x^4)^2-10(x^4)+9 = 0#
We can factor this to find:
#0 = (x^4)^2-10(x^4)+9 = (x^4-1)(x^4-9)#
Each of the remaining quartic factors is a difference of squares, so we can use:
#A^2-B^2 = (A-B)(A+B)#
to find:
#x^4-1 = (x^2)^2-1^2 = (x^2-1)(x^2+1)#
#x^4-9 = (x^2)^2 - 3^2 = (x^2-3)(x^2+3)#
The remaining quadratic factors will all factor as differences of squares too, but we need to use irrational and/or complex coefficients to do some of them::
#x^2-1 = x^2-1^2 = (x-1)(x+1)#
#x^2+1 = x^2-i^2 = (x-i)(x+i)#
#x^2-3 = x^2-(sqrt(3))^2 = (x-sqrt(3))(x+sqrt(3))#
#x^2+3 = x^2-(sqrt(3)i)^2 = (x-sqrt(3)i)(x+sqrt(3)i)#
Hence the zeros of the original octic polynomial are:
#x = +-1, +-i, +-sqrt(3), +-sqrt(3)i#
