Question

How do you factor `2x^9-2`?

Answer 1

`2x^9-2`

`= 2(x^9 - 1)`

`= 2((x^3)^3 - 1^3)`

We know that `color(blue)(a^3 - b^3 = (a - b)(a^2 + ab + b^2)`

`= 2(x^3 - 1)((x^3)^2 + (x^3)(1) + 1^2)`

`= 2(x^3 - 1)(x^6 + x^3 + 1)`

`= 2(x^3 - 1^3)(x^6 + x^3 + 1)`

Applying the Difference of Cubes Formula again, we get

`= 2(x - 1)(x^2 + (x)(1) + 1^2)(x^6 + x^3 + 1)`

`color(green)( = 2(x - 1)(x^2 + x + 1)(x^6 + x^3 + 1)`

As none of the Factors can be factorised further, the above becomes the factorised form of `2x^9-2`