How do you factor #x^16-81#?

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Answer 1 · 2017-07-14T13:55:13

#(x - x_0)(x - x_1)(x - x_2) ... (x - x_15)#

When we solve #z^16 = 81#, we find 16 roots.

#x_n = 81^(1/16) (cos frac{2 pi * n}{16} + i sin frac{2 pi * n}{16})#

#x_n = 3^(1/4) (cos frac{pi * n}{8} + i sin frac{pi * n}{8})#

Answer 2 · 2017-07-14T13:55:33

As the difference of two squares #(x^8 -9) xx (x^8 +9) #

# x^16# can be consider the square of # x^8 xx x^8#

81 can be consider the square of # 9xx 9#

when the binomials of two squares are multiplied as alternatively positive and negative terms the middle term falls out.

#( x^8 -9) xx (x^8 + 9 ) = x^16 - 9x^8 + 9x^8 - 81#

# -9x^8 + 9x^8 = 0# so

# (x^8 -9) xx (x^8 +9) = x^16 -81#

Answer 3 · 2017-07-14T15:00:40

#(x^8+9)((x^4+3)(x^4-3)#

#x^16-81#

using difference of squares

#=(x^8+9)(x^8-9)#

using DoS on the second bracket

#(x^8+9)((x^4+3)(x^4-3)#