How do you factor #2n^3 +6n^2 +10n#?

1 Answer
Sep 23, 2016

Answer:

#2n^3+6n^2+10n = 2n(n^2+3n+5)#

#color(white)(2n^3+6n^2+10n) = 2n(n+3/2-sqrt(11)/2i)(n+3/2+sqrt(11)/2i)#

Explanation:

Given:

#2n^3+6n^2+10n#

Note that all of the terms are divisible by #2n#, so we can separate that out as a factor:

#2n^3+6n^2+10n = 2n(n^2+3n+5)#

Looking at the remaining quadratic in #n# we find:

#n^2+3n+5 = n^2+3n+9/4+11/4#

#color(white)(n^2+3n+5) = (n+3/2)^2+11/4#

For any Real value of #n# this will be positive, hence #n^2+3n+5# has no linear factors with Real coefficients.

We can factor it with Complex coefficients, which can be done using the difference of squares identity:

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

with #a=(n+3/2)# and #b=sqrt(11)/2i# as follows:

#n^2+3n+5 = (n+3/2)^2+11/4#

#color(white)(n^2+3n+5) = (n+3/2)^2-(sqrt(11)/2i)^2#

#color(white)(n^2+3n+5) = ((n+3/2)-sqrt(11)/2i)((n+3/2)+sqrt(11)/2i)#

#color(white)(n^2+3n+5) = (n+3/2-sqrt(11)/2i)(n+3/2+sqrt(11)/2i)#

where #i# is the imaginary unit, which satisfies #i^2=-1#

Hence:

#2n^3+6n^2+10n = 2n(n+3/2-sqrt(11)/2i)(n+3/2+sqrt(11)/2i)#