How do you factor $x^{3} + 3 x^{2} - 2 x - 6$ $x^3+3x^2-2x-6$ ?

Question

15322 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-09T18:32:55

$(x + 3) (x + \sqrt{2}) (x - \sqrt{2})$ $(x + 3)(x + sqrt(2))(x - sqrt(2))$

We have: $x^{3} + 3 x^{2} - 2 x - 6$ $x^(3) + 3 x^(2) - 2 x - 6$

$= (x^{3} + 3 x^{2}) + (- 2 x - 6)$ $= (x^(3) + 3 x^(2)) + (- 2 x - 6)$

Let's take $x^{2}$ $x^(2)$ in common in the first set of parentheses and $- 2$ $- 2$ in common in the second set of parentheses, respectively:

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

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

Then, let's take $x + 3$ $x + 3$ in common:

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

If you allow surds to be used, then the second set of parentheses can be factorised further:

$= (x + 3) (x + \sqrt{2}) (x - \sqrt{2})$ $= (x + 3)(x + sqrt(2))(x - sqrt(2))$

How do you factor x3+3x2−2x−6x^3+3x^2-2x-6?