n^3+n^2-3n-3

Question

Factorise completely 

Answer 1

In order to factorize the polynomial I'm going to use the method grouping. Basically this means I'm going to factor the first two terms (find what they have in common) and factor the last two terms (find what they have in common).

Factor first two terms:  n^3+n^2. They both have a n^2 in commons so I'll take that out and: n^3+n^2=n^2(n+1)

Factor last two terms: -3n-3. They both have an -3 in common so I'll take that out and:
 -3n-3=-3(n+1). 

Now my polynomials looks like this: n^2(n+1)-3(n+1). Notice that both the first terms n^2(n+1) and the second term -3(n+1) have a (n+1) in common. I can take that out exactally like I did before hand. So it becomes: (n^2-3)(n+1). You can also factor the (n^2-3) more since it follows the pattern Difference of Squares rule (http://www.mathwords.com/f/factoring_rules.htm). So n^2-3=(n-square root of 3)(n+squre root of 3)

Here is the math in completion:

n^3+n^2-3n-3=n^2(n+1)-3(n+1)=(n^2-3)(n+1)=(n-square root of 3)(n+square of 3)(n+1).