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

3 Answers
Jul 14, 2017

Answer:

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

Explanation:

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})#

Jul 14, 2017

Answer:

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

Explanation:

# 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#

Jul 14, 2017

Answer:

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

Explanation:

#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)#