How do you factor #14r^4-378rs^3#?

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Answer 1 · 2016-05-09T19:18:57

#14r^4-378rs^3 = 14r(r-3s)(r^2+3rs+9s^2)#

First note that both terms are divisible by #14r#, so separate that out as a factor first.

#14r^4-378rs^3 = 14r(r^3-27s^3)#

Next note that both #r^3# and #27s^3=(3s)^3# are perfect cubes, so we can use the difference of cubes identity:

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

with #a=r# and #b=3s# as follows:

#r^3-27s^3#

#=r^3-(3s)^3#

#=(r-3s)(r^2+r(3s)+(3s)^2)#

#=(r-3s)(r^2-3rs+9s^2)#

Putting it all together:

#14r^4-378rs^3 = 14r(r-3s)(r^2+3rs+9s^2)#