Question

How do I factor the following expression?

Answer 1

$\bf x^6-9x^4-81x^2+729\implies \stackrel{\textit{let's do some grouping}}{(x^6-9x^4)-(81x^2-729)} \\\\\\ \stackrel{\textit{some common factoring}}{x^4\boxed{(x^2-9)}-81\boxed{(x^2-9)}}\implies \stackrel{\textit{common factoring}}{\boxed{(x^2-9)}(x^4-81)} \\\\\\ \stackrel{\textit{difference of squares}}{[x^2-3^2][(x^2)^2-(3^2)^2]}\implies (x-3)(x+3)\stackrel{\textit{difference of squares}}{(x^2-3^2)}(x^2+3^2) \\\\\\ (x-3)(x+3)~(x-3)(x+3)~(x^2+9)$

Answer 2

On a hunch, I decided to check whether 729 is a 6th power, and found that it is:

729^(1/6) = 3.

Next, I decided to divide x-3 into x^6 + 0x^5 - 9x^4 + 0x^3 - 81x^2 + 0x + 729, through synthetic division and using 3 as my divisor. This left no remainder. The coefficients of the quotient were as follows: 3 3 0 0 -81 -243, which represents:

3x^5 + 3x^4 + 0x^3 + 0x^2 -81x -243

Next, I applied "factoring by grouping:"

3x^4(x+1) -81(x+3) = (x+3)(3x^4 - 81).

Note that 3x^4 - 81 is the same as 3(x^4-27).

Thus, the original polynomial factors into (x-3)(3)(x^4 - 27).