# How do you factor  6x^3+48?

Dec 18, 2015

Separate out the scalar factor $6$ then use the sum of cubes identity to find:

$6 {x}^{3} + 48 = 6 \left(x + 2\right) \left({x}^{2} - 2 x + 4\right)$

#### Explanation:

First separate out the common scalar factor $6$ to find:

$6 {x}^{3} + 48 = 6 \left({x}^{3} + 8\right)$

Then notice that both ${x}^{3}$ and $8 = {2}^{3}$ are perfect cubes, so work well with the sum of cubes identity:

${A}^{3} + {B}^{3} = \left(A + B\right) \left({A}^{2} - A B + {B}^{2}\right)$

With $A = x$ and $B = 2$ we find:

${x}^{3} + 8 = \left({x}^{2} + {2}^{3}\right) = \left(x + 2\right) \left({x}^{2} - 2 x + 4\right)$

Putting it together we get:

$6 {x}^{3} + 48 = 6 \left(x + 2\right) \left({x}^{2} - 2 x + 4\right)$

This has no simpler factors with Real coefficients, as you can check by looking at the discriminant $\Delta$ of $\left({x}^{2} - 2 x + 4\right)$

$\Delta = {b}^{2} - 4 a c = {\left(- 2\right)}^{2} - \left(4 \times 1 \times 4\right) = 4 - 16 = - 12$

Since $\Delta < 0$ this quadratic has no Real zeros and no linear factors with Real coefficients.

Dec 18, 2015

First factor out the 6 ...

#### Explanation:

$6 \left({x}^{3} + 8\right)$

Now use the identity for the sum of cubes ...

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

$6 \left(x + 2\right) \left({x}^{2} - 2 x + 4\right)$

hope that helped