How do you factor $16 - 2 x^{3}$ $16 - 2x^3$ ?

Question

3508 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-08T20:27:59

The fully factored polynomial is $= - 2 (x - 2) (x^{2} + 2 x + 4)$ $=-2(x-2)(x^2+2x+4)$ .

First, factor out the $2$ $2$ :

$= 16 - 2 x^{3}$ $color(white)=16-2x^3$

$= 2 (8 - x^{3})$ $=2(8-x^3)$

Then, use the difference of cubes factoring:

$= 2 (2^{3} - x^{3})$ $=2(2^3-x^3)$

$= 2 (2 - x) (2^{2} - 2 x + x^{2})$ $=2(2-x)(2^2-2x+x^2)$

$= 2 (2 - x) (4 - 2 x + x^{2})$ $=2(2-x)(4-2x+x^2)$

Lastly, reorder the factors so that $x$ $x$ is in the front of each one:

$= 2 (2 - x) (x^{2} + 2 x + 4)$ $=2(2-x)(x^2+2x+4)$

$= 2 (- x + 2) (x^{2} + 2 x + 4)$ $=2(-x+2)(x^2+2x+4)$

$= 2 (- 1 \cdot (x - 2)) (x^{2} + 2 x + 4)$ $=2(-1*(x-2))(x^2+2x+4)$

$= - 2 (x - 2) (x^{2} + 2 x + 4)$ $=-2(x-2)(x^2+2x+4)$

That's as factored as it gets. Hope this helped!

How do you factor 16−2x316 - 2x^3?