I know answer; am struggling to explain how to get it: 2n^3+n^2-11n+12/n+3

[marykateh]

2n^3+n^2-11n+12/n+3

So the objective is to cancel n+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

 

[sinx]

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

(n+3)(2n^2-5n+4)/(n+3)
n+3 "cancels out" (n+3 divided by itself equals 1).

 

[Harry_the_cat]

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

Specifically what numbers do you mean?

 

[JeffM]

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.

 

[Denis]

2n^3+n^2-11n+12/n+3

Brackets required:
(2n^3+n^2-11n+12)/(n+3) ; ok?

Have you tried long division?

 

[mmm4444bot]

… algebraic division. UGH!!!! …

I'm thinking the UGH comes from writing out multiple, longhand division steps completely. For readers who may not be familiar with it, the synthetic form of polynomial division (aka: Synthetic Division) provides a nice shortcut.

[FONT=Courier New]

-3    2   1  -11   12
         -6   15  -12
    -----------------
      2  -5    4    0
[/FONT]