How do you factor x^16-81?

3 Answers
Jul 14, 2017

$\left(x - {x}_{0}\right) \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \ldots \left(x - {x}_{15}\right)$

Explanation:

When we solve ${z}^{16} = 81$, we find 16 roots.

${x}_{n} = {81}^{\frac{1}{16}} \left(\cos \frac{2 \pi \cdot n}{16} + i \sin \frac{2 \pi \cdot n}{16}\right)$

${x}_{n} = {3}^{\frac{1}{4}} \left(\cos \frac{\pi \cdot n}{8} + i \sin \frac{\pi \cdot n}{8}\right)$

Jul 14, 2017

As the difference of two squares $\left({x}^{8} - 9\right) \times \left({x}^{8} + 9\right)$

Explanation:

${x}^{16}$ can be consider the square of ${x}^{8} \times {x}^{8}$

81 can be consider the square of $9 \times 9$

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

$\left({x}^{8} - 9\right) \times \left({x}^{8} + 9\right) = {x}^{16} - 9 {x}^{8} + 9 {x}^{8} - 81$

$- 9 {x}^{8} + 9 {x}^{8} = 0$ so

$\left({x}^{8} - 9\right) \times \left({x}^{8} + 9\right) = {x}^{16} - 81$

Jul 14, 2017

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

Explanation:

${x}^{16} - 81$

using difference of squares

$= \left({x}^{8} + 9\right) \left({x}^{8} - 9\right)$

using DoS on the second bracket

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