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

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Answer 1 · 2015-04-17T05:38:04

#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#