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

Question

Julia Adams · Accepted Answer

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.

Eli Assouline · Answer

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

Jameson Allen · Answer

undefined To factor the expression 6x^3 + 48, we first look for the greatest common factor, which is 6. So, we can factor out 6 from both terms:
6(x^3 + 8).
Now, we observe that x^3 + 8 can be factored further using the sum of cubes formula:
x^3 + 8 = (x + 2)(x^2 - 2x + 4).
So, the fully factored expression is:
6(x + 2)(x^2 - 2x + 4).