I know what the answer is (r^2+6r+4), but I don't know how I got it! Can someone help, please. Original Problem: (r^3+5r^2-2r-4)/(r-1)

We are asked to perform the operation `(r^3+5r^2-2r-4 )/(r-1) ` :

Since the denominator is a monic (leading coefficient 1) linear function, we can easily use synthetic division:

1  |  1   5   -2   -4
    -----------------
       1   6    4     0

Reading the quotient and remainder from the bottom row we get `r^2+6r+4 ` with remainder zero.

If you are unfamiliar with synthetic division, or the problem has a nonlinear divisor, you can use long division:

      r^2   +   6r   +   4
      --------------------
r-1| r^3  +5r^2  -2r  -4
      r^3   -r^2
      -----------
               6r^2  -2r
               6r^2  -6r
               ---------
                         4r  -4
                         4r  -4
                         ------
                               0

Finally, if you do not know either division method you can factor the numerator and divide out any common factors:

`(r^3+5r^2-2r-4)/(r-1)=((r-1)(r^2+6r+4))/(r-1)=r^2+6r+4 `

One way to factor the numerator:

`r^3+5r^2-2r-4=(r^3-1)+5r^2-2r-4+1 `  add/subtract 1

`=(r-1)(r^2+r+1)+5r^2-2r-3 `  Factor the difference of cubes

`=(r-1)(r^2+r+1)+(r-1)(5r+3) ` Factor the quadratic term

`=(r-1)(r^2+r+1+5r+3)=(r-1)(r^2+6r+4) ` using the distributive property.