How do you factor $3x^3-24$ ?

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Answer 1 · 2016-02-14T05:39:24

$3x^3-24=3(x-2)(x^2+2x+4)$
$= 3(x-2)(x+1+sqrt(3)i)(x+i-sqrt(3)i)$

Using the difference of cubes formula $a^3 - b^3 = (a-b)(a^2+ab+b^2)$
we have

$3x^3-24 = 3(x^3-8)$

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

$=3(x-2)(x^2+2x+4)$

If we only allow real numbers, we are done. If we allow complex numbers, we can use the quadratic formula to factor $x^2+2x+4$ by finding its roots.

$x^2+2x+4 = 0$
$<=> x = (-2+-sqrt(2^2-4(1)(4)))/(2(1))$

$=(-2+-sqrt(-12))/2$

$=-1+-sqrt(-3)$

$=-1+-sqrt(3)i$

Thus

$3x^3-24 = 3(x-2)(x+1+sqrt(3)i)(x+i-sqrt(3)i)$

How do you factor 3x^3-243x^3-24?