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

##### 2 Answers

#### Answer:

Separate out the scalar factor

#6x^3+48 = 6(x+2)(x^2-2x+4)#

#### Explanation:

First separate out the common scalar factor

#6x^3+48 = 6(x^3+8)#

Then notice that both

#A^3+B^3 = (A+B)(A^2-AB+B^2)#

With

#x^3+8 = (x^2+2^3) = (x+2)(x^2-2x+4)#

Putting it together we get:

#6x^3+48 = 6(x+2)(x^2-2x+4)#

This has no simpler factors with Real coefficients, as you can check by looking at the discriminant

#Delta = b^2-4ac = (-2)^2-(4xx1xx4) = 4-16 = -12#

Since

#### Answer:

First factor out the 6 ...

#### Explanation:

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

hope that helped