Example 2
Completion requirements
| Example 2 |
|
Simplify the expression \(\frac{{c + 2}}{{c^2 - c - 20}} - \frac{{3c^2 - 14c - 5}}{{2c^2 + 7c + 6}} \div \frac{{c^2 - c - 20}}{{2c + 3}}\). Identify any non-permissible values.
Step 1: Factor all polynomials to determine the non-permissible values.
NPVs:
Step 2: Determine which operation to perform first, and simplify.
According to BEDMAS, division comes before subtraction.
Pause, and identify any new NPVs:
There are no new NPVs.
Continue simplifying.
Step 3: Perform the last operation, and simplify.
Subtraction is the last operation.
\[\frac{{c + 2}}{{c^2 - c - 20}} - \frac{{3c^2 - 14c - 5}}{{2c^2 + 7c +
6}} \div \frac{{c^2 - c - 20}}{{2c + 3}} = \frac{{c + 2}}{{\left( {c -
5} \right)\left( {c + 4} \right)}} - \frac{{\left( {3c + 1}
\right)\left( {c - 5} \right)}}{{\left( {2c + 3} \right)\left( {c + 2}
\right)}} \div \frac{{\left( {c - 5} \right)\left( {c + 4} \right)}}{{2c
+ 3}}\]
NPVs:
\(\begin{align}
c - 5 &\ne 0 \\
c &\ne 5
\end{align}\)
c - 5 &\ne 0 \\
c &\ne 5
\end{align}\)
\(\begin{align}
c + 4 &\ne 0 \\
c &\ne - 4
\end{align}\)
c + 4 &\ne 0 \\
c &\ne - 4
\end{align}\)
\(\begin{align}
2c + 3 &\ne 0 \\
c &\ne -\frac{3}{2}
\end{align}\)
2c + 3 &\ne 0 \\
c &\ne -\frac{3}{2}
\end{align}\)
\(\begin{align}
c + 2 &\ne 0 \\
c &\ne - 2
\end{align}\)
c + 2 &\ne 0 \\
c &\ne - 2
\end{align}\)
Step 2: Determine which operation to perform first, and simplify.
According to BEDMAS, division comes before subtraction.
\[\frac{{c + 2}}{{\left( {c - 5} \right)\left( {c + 4} \right)}} -
{\color{red}\frac{{\left( {3c + 1} \right)\left( {c - 5} \right)}}{{\left( {2c + 3}
\right)\left( {c + 2} \right)}} \div \frac{{\left( {c - 5} \right)\left(
{c + 4} \right)}}{{2c + 3}}} = \frac{{c + 2}}{{\left( {c - 5}
\right)\left( {c + 4} \right)}} - {\color{red}\frac{{\left( {3c + 1} \right)\left(
{c - 5} \right)}}{{\left( {2c + 3} \right)\left( {c + 2} \right)}} \cdot \frac{2c + 3}{\left( {c - 5} \right)\left( {c + 4} \right)}}\]
Pause, and identify any new NPVs:
There are no new NPVs.
Continue simplifying.
\[\begin{align}
\frac{{c + 2}}{{\left( {c - 5} \right)\left( {c + 4} \right)}} - {\color{red}\frac{{\left( {3c + 1} \right) \cancel {\color{#444}\left( {c - 5} \right)}}}{{\cancel {\color{#444}\left( {2c + 3} \right)}\left( {c + 2} \right)}}} \cdot \frac{{{\color{red} \cancel {\color{#444}\left( {2c + 3} \right)}}}}{{{\color{red} \cancel {\color{#444}\left( {c - 5} \right)}\left( {c + 4} \right)}}} &= \frac{{c + 2}}{{\left( {c - 5} \right)\left( {c + 4} \right)}} - {\color{red}\frac{{3c + 1}}{{c + 2}} \cdot \frac{1}{{c + 4}}} \\
&= \frac{{c + 2}}{{\left( {c - 5} \right)\left( {c + 4} \right)}} - {\color{red}\frac{{3c + 1}}{{\left( {c + 2} \right)\left( {c + 4} \right)}}} \\
\end{align}\]
\frac{{c + 2}}{{\left( {c - 5} \right)\left( {c + 4} \right)}} - {\color{red}\frac{{\left( {3c + 1} \right) \cancel {\color{#444}\left( {c - 5} \right)}}}{{\cancel {\color{#444}\left( {2c + 3} \right)}\left( {c + 2} \right)}}} \cdot \frac{{{\color{red} \cancel {\color{#444}\left( {2c + 3} \right)}}}}{{{\color{red} \cancel {\color{#444}\left( {c - 5} \right)}\left( {c + 4} \right)}}} &= \frac{{c + 2}}{{\left( {c - 5} \right)\left( {c + 4} \right)}} - {\color{red}\frac{{3c + 1}}{{c + 2}} \cdot \frac{1}{{c + 4}}} \\
&= \frac{{c + 2}}{{\left( {c - 5} \right)\left( {c + 4} \right)}} - {\color{red}\frac{{3c + 1}}{{\left( {c + 2} \right)\left( {c + 4} \right)}}} \\
\end{align}\]
Step 3: Perform the last operation, and simplify.
Subtraction is the last operation.
\[\begin{align}
\frac{{\left( {c + 2} \right) {\color{red}\left( {c + 2} \right)}}}{{\left( {c - 5} \right)\left( {c + 4} \right) {\color{red}\left( {c + 2} \right)}}} - \frac{{\left( {3c + 1} \right) {\color{red}\left( {c - 5} \right)}}}{{\left( {c + 2} \right)\left( {c + 4} \right) {\color{red}\left( {c - 5} \right)}}} &= \frac{{\left( {c + 2} \right)^2 }}{{\left( {c - 5} \right)\left( {c + 4} \right)\left( {c + 2} \right)}} - \frac{{\left( {3c + 1} \right)\left( {c - 5} \right)}}{{\left( {c - 5} \right)\left( {c + 4} \right)\left( {c + 2} \right)}} \\
&= \frac{{c^2 + 4c + 4}}{{\left( {c - 5} \right)\left( {c + 4} \right)\left( {c + 2} \right)}} - \frac{{3c^2 - 14c - 5}}{{\left( {c - 5} \right)\left( {c + 4} \right)\left( {c + 2} \right)}} \\
&= \frac{{c^2 + 4c + 4 - \left( {3c^2 - 14c - 5} \right)}}{{\left( {c - 5} \right)\left( {c + 4} \right)\left( {c + 2} \right)}} \\
&= \frac{{ - 2c^2 + 18c + 9}}{{\left( {c - 5} \right)\left( {c + 4} \right)\left( {c + 2} \right)}},\thinspace c \ne - 4,\thinspace - 2, \thinspace -\frac{3}{2}, \thinspace 5 \\
\end{align}\]
\frac{{\left( {c + 2} \right) {\color{red}\left( {c + 2} \right)}}}{{\left( {c - 5} \right)\left( {c + 4} \right) {\color{red}\left( {c + 2} \right)}}} - \frac{{\left( {3c + 1} \right) {\color{red}\left( {c - 5} \right)}}}{{\left( {c + 2} \right)\left( {c + 4} \right) {\color{red}\left( {c - 5} \right)}}} &= \frac{{\left( {c + 2} \right)^2 }}{{\left( {c - 5} \right)\left( {c + 4} \right)\left( {c + 2} \right)}} - \frac{{\left( {3c + 1} \right)\left( {c - 5} \right)}}{{\left( {c - 5} \right)\left( {c + 4} \right)\left( {c + 2} \right)}} \\
&= \frac{{c^2 + 4c + 4}}{{\left( {c - 5} \right)\left( {c + 4} \right)\left( {c + 2} \right)}} - \frac{{3c^2 - 14c - 5}}{{\left( {c - 5} \right)\left( {c + 4} \right)\left( {c + 2} \right)}} \\
&= \frac{{c^2 + 4c + 4 - \left( {3c^2 - 14c - 5} \right)}}{{\left( {c - 5} \right)\left( {c + 4} \right)\left( {c + 2} \right)}} \\
&= \frac{{ - 2c^2 + 18c + 9}}{{\left( {c - 5} \right)\left( {c + 4} \right)\left( {c + 2} \right)}},\thinspace c \ne - 4,\thinspace - 2, \thinspace -\frac{3}{2}, \thinspace 5 \\
\end{align}\]