Functions: Fractional functions
Long division with polynomials
If the degree of the numerator of a quotient function is greater than or equal to the degree of the denominator, we can write the quotient function as the sum of a quotient and a rest. The quotient is a polynomial and the remainder is again a quotient function.
Every function of the form
\[f(x)=\frac{\blue{p(x)}}{\orange{q(x)}}\]
where #\blue p# and #\orange q# are polynomials and #\text{degree }\blue{p(x)} \geq \text{degree } \orange{q(x)}#,
can be rewritten using long division in the form
\[f(x) = \green{s(x)}+\frac{\purple{r(x)}}{\orange{q(x)}}\]
where #\green s# is a polynomial and #\text{degree }\purple{r(x)} < \text{degree }\orange{q(x)}#.
Example
\[ f(x) = \frac{\blue{2x^2+5x+5}}{\orange{x+1}} \]
gives
\[f(x) = \green{2x+3} + \dfrac{\purple{2}}{\orange{x+1}}\]
We use the following step-by-step guide for the division of polynomials using long division.
Long division
|
Step-by-step |
Example |
|
|
For the quotient function we use long division \[f(x)=\frac{\blue{p(x)}}{\orange{q(x)}}\] |
\[f(x)=\frac{\blue{2x^2+5x+5}}{\orange{x+1}}\] |
|
| Step 1 |
Compose the long division as follows: #\require{enclose} |
\[ \require{enclose} |
| Step 2 |
Now find an #\green{\text{expression}}# such that if it is multiplied by #\orange{q(x)}#, the term with the highest degree is equal to the term with the highest degree of #\blue{p(x)}#. In the example on the right, we chose #\green{2x}# because #\green{2x} \cdot (\orange{x+1})# is equal to #2x^2+2x#. We put this term #\green{2x}# above the line in the long division. Now subtract the obtained expression from #\blue{p(x)}# to get an expression #\purple{r(x)}#. In the example, we found #\purple{3x+5}#. Repeat this process until #\text{degree }\purple{r(x)} < \text{degree }\orange{q(x)}#. |
# \require{enclose} |
| Step 3 | It now follows that \[ \begin{array}{rcl} f(x) &=& \dfrac{\blue{p(x)}}{\orange{q(x)}} \\
|
\[\begin{array}{rcl} f(x) &=&\dfrac{\blue{2x^2+5x+5}}{\orange{x+1}} \\ |
| Step 1 |
We first compose the long division: \[\require{enclose} |
| Step 2 |
Long division gives \[ \require{enclose} |
| Step 3 |
According to the theory, it now follows that \[\begin{array}{rcl} f(x) &=&\dfrac{{6x+6}}{{3x+7}} \\
|
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