How do you factor #n^3 + 4n^2 - 21n = 0#?

2 Answers
May 10, 2018

Answer:

#n(n-3)(n+7)#

Explanation:

Given: #n^3+4n^2-21n=0#.

Factor out #n# first.

#n(n^2+4n-21)=0#.

Factor #(n^2+4n-21)#.

#=>n(n^2+7n-3n-21)#

#=>n(n(n+7)-3(n+7))#

#=>n(n-3)(n+7)#

May 10, 2018

Answer:

#n(n-3)(n+7)=0#

Explanation:

Notice that on the LHS we have #n# in each term. So we can factor that out leaving a quadratic.

#n(n^2+4x-21)=0#

Notice that #3xx7=21# and that #7-3=4# So we can use this in the form of:

#n(n-3)(n+7)=0 color(white)("d")larr# this is now factorised as required.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The very fact that this is presented as an equation implies that you are required to determine feasible values for n that satisfy the given condition.

#n=0;color(white)("..d") n=+3;color(white)("..d") n=-7#