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

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Answer 1 · 2015-12-18T19:45:06

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

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

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

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

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

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

#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# of #(x^2-2x+4)#

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

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

Answer 2 · 2015-12-18T19:47:17

First factor out the 6 ...

#6(x^3+8)#

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

#a^3+b^3=(a+b)(a^2-ab+b^2)#

#6(x+2)(x^2-2x+4)#

hope that helped