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)

Saturday, March 16, 2013

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 $\displaystyle\frac{{{{r}}^{{3}}+{5}{{r}}^{{2}}-{2}{r}-{4}}}{{{r}-{1}}}$ :

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 $\displaystyle{{r}}^{{2}}+{6}{r}+{4}$ 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:

$\displaystyle\frac{{{{r}}^{{3}}+{5}{{r}}^{{2}}-{2}{r}-{4}}}{{{r}-{1}}}=\frac{{{\left({r}-{1}\right)}{\left({{r}}^{{2}}+{6}{r}+{4}\right)}}}{{{r}-{1}}}={{r}}^{{2}}+{6}{r}+{4}$

One way to factor the numerator:

$\displaystyle{{r}}^{{3}}+{5}{{r}}^{{2}}-{2}{r}-{4}={\left({{r}}^{{3}}-{1}\right)}+{5}{{r}}^{{2}}-{2}{r}-{4}+{1}$ add/subtract 1

$\displaystyle={\left({r}-{1}\right)}{\left({{r}}^{{2}}+{r}+{1}\right)}+{5}{{r}}^{{2}}-{2}{r}-{3}$ Factor the difference of cubes

$\displaystyle={\left({r}-{1}\right)}{\left({{r}}^{{2}}+{r}+{1}\right)}+{\left({r}-{1}\right)}{\left({5}{r}+{3}\right)}$ Factor the quadratic term

$\displaystyle={\left({r}-{1}\right)}{\left({{r}}^{{2}}+{r}+{1}+{5}{r}+{3}\right)}={\left({r}-{1}\right)}{\left({{r}}^{{2}}+{6}{r}+{4}\right)}$ using the distributive property.

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Saturday, March 16, 2013