You should have had this in high school algebra or college algebra, but just in case you're rusty, here's how we did the division in problem 4 of Practice Problems 4. We have an advantage in that we know what our final expression should look like; we know what the factors should be. Solving a cubic equation when you don't know the factors is possible, but more difficult.

Our polynomial is: 2n³ + 9n² + 13n + 6.
We know that the factors should be n + 1, n + 2, and 2n + 3. In this case, we can simply test out our factors directly. If (n+1)(n+2)(2n+3) = 2n³ + 9n² + 13n + 6 (and it does), we're done.

But we won't always know the factors in advance. We may suspect that there's an n+1 lurking in there somewhere, but not be sure. What then?

Work it one step at a time, basing your value for the quotient on the highest power of n in the expression you're dealing with. Then bring terms down from the dividend in order to process the next term.
 

                 2n²  + 7n +   6
               ___________________
     n+1 )    2n³ + 9n² + 13n + 6
                 2n³ + 2n²
                ------------
                         7n²  + 13n
                         7n²  + 7n
                        ------------
                                   6n + 6
                                   6n + 6
                                  ---------

So   2n³ + 9n² + 13n + 6 = (n+1)(2n² + 7n + 6).  Factoring the quadratic expression should be a cakewalk, but here's the worked example just in case:

        __2n_+__3____
n+2 ) 2n² + 7n + 6
         2n² + 4n
        ----------
                  3n + 6
                  3n + 6
                 --------

Practice Problems 4 Answer Key

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