How do you solve the equation #n^3+2n^2-35n=0#?

2 Answers
Feb 5, 2015

Okay so the first step will be removing the gcf

So lets write this as # n (n^2 + 2n- 35)#
so lets faCtor# (n^2 + 2n- 35)#

this CAN BE REWRITTEN AS # n(n+7)(n-5)#

Now using the 0 theorem we can say that one of the 3 polynomials which are being multiplied have to be = 0

So hence the set of values for n are #(0,-7,5)#

Feb 5, 2015

You can take one #n# out and have a quadratic equation.

#n^3+2n^2-35n=n*(n^2+2n-35)#

Now we allready have one possible solution: #n=0#

Let's look at what we have left. A quadratic equation of the form
#ax^2+bx+c=0# where #a=1#, #b=2# and #c=-35#

Now we have to find two numbers that, when multiplied, give a product of #-35# and when added/subtracted give a sum of #2#

These would be #7# and #-5#
So the quadratic part factors into #(n+7)(n-5)#

And the whole original equation will factor into:

#n^3+2n^2-35n=n*(n+7)(n-5)=0#

Answer :
#n=0orn=-7orn=5#