2n^3+n^2-11n+12/n+3
So the objective is to cancel x+3 from the top and the bottom.
(n+3)(2n^2-5n+4)/n+3
Final simplification = 2n^2-5n+4
I need an explanation of how we know to use these numbers. Any suggestions are welcome.
Thanks
-Mk
Please read
https://www.freemathhelp.com/forum/threads/112086-Guidelines-Summary
I am not sure that I understand your question, but I think you are asking how to find the factoring that allows you to cancel the denominator. The first point to realize is that there may not be a factoring that permits cancellation.
There are basically two ways to find such a factoring if it exists. One way is to factor the numerator using the clue that you want the denominator to be a factor. (The other way is algebraic division. UGH!!!!)
\(\displaystyle 2n^3 + n^2 - 11n + 12 = (n + 3)(an^2 + bn + c)\)
If such a factoring exists, n times an^2 must equal 2n^3. Therefore a must equal 2. Similarly, 3 times c must equal 12. Therefore, c must equal 4. So
\(\displaystyle 2n^3 + n^2 - 11n + 12 = (n + 3)(2n^2 + bn + 4).\)
Now, if the desired factoring exists, bn^2 + 6n^2 must equal n^2. Thus, b must equal - 5. But remember: there may not be such a factoring. So we
MUST CHECK.
\(\displaystyle (n + 3)(2n^2 - 5n + 4) = 2n^3 - 5n^2 + 4n + 6n^2 - 15n + 12 =\)
\(\displaystyle 2n^3 + 6n^2 - 5n^2 + 4n - 15n + 12 = 2n^3 + n^2 - 11n + 12.\)
So the factoring works.