Factorize: (n^2 + 2n +3 ) / (2n^3 + 5n^2 + 8n +3)

[jetoso]

I have a really bad time thinking on this:

Factorize:

(n^2 + 2n +3 ) / (2n^3 + 5n^2 + 8n +3) such that we end up with 1/(2n +1).

I might be forgetting how to complete the square or the cube or something but I can not find a way to reduce it.

Any advice?

 

[berkeman]

You should be able to factor the numerator. Do you know how to set up the two simultaneous equations to figure out the a and b in (n+a)(n+b) = n^2+[a+b]n+[a*b] ?

 

[jetoso]

Answer

Say for n^2 + 2n + 3 we can write it as (n+1)^2 + 2, so we have that (n+1)(n+1) + 2 = n^2 + 2n + 3

 

[HallsofIvy]

You are told that you must wind up with an extra factor of 2x+1 in the denominator. Is it that hard to divide n² + 2n +3 and 2n³ + 5n² + 8n +3 by 2n +1?

 

[berkeman]

Say for n^2 + 2n + 3 we can write it as (n+1)^2 + 2, so we have that (n+1)(n+1) + 2 = n^2 + 2n + 3

LOL, that's so much better a way to start. I suggested the brute strength way to start, but spaced the alternate constant form. Good stuff jetoso.

 

[jetoso]

I got it

Well, I think I did not pay too much attention to this problem, it is very straight forward by the way, look:

2n^3 + 5n^2 + 8n + 3 = (2n + 1) (n^2 + 2n + 3)

Thus, since the numerator is n^2 + 2n + 3, then it follows that:

(n^2 + 2n + 3)/(2n + 1) (n^2 + 2n + 3) = 1/(2n + 1)


Sorry to bother you about this.
Thanks!