Factor #3n^4+21n^3+27n^2#?

Question

1575 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-25T15:52:06

#3n^2(n^2+7n+9)#

Expanding we get
#3n^4+21n^3+27n^2#

Answer 2 · 2018-05-25T16:04:12

#n= {0,7/2+sqrt13/2,-7/2-sqrt13/2}#

Factor:

#3n^4+21n^3+27n^2 =0#

First solve by using the distributive property this expression has a GCF of #3n^2#

#3n^2(n^2+7n+9) = 0#

We now know one factor #3n^2 =0#, so #n=0#

To factor #n^2+7n+9# we will have to use the quadratic formula:

#n = (-b+-sqrt(b^2-4ac))/(2a)#

#a=1#
#b=7#
#c=9#

#n = (-7+-sqrt(7^2-4*1*9))/(2*1)#

#n = -7/2+-sqrt13/2#

so all factors are:

#n= {0,7/2+sqrt13/2,-7/2-sqrt13/2}#