Long division with polynomials

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 remainder. 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

\[f(x) = \green{2x+3} + \dfrac{\purple{2}}{\orange{x+1}}\]

Remainder is zero

If the remainder #\purple{r(x)}# is zero, the quotient function #f(x)# has been simplified to the polynomial #\green{s(x)}#. But be aware, the domain of #f(x)# contains all #x# for which #\orange{q(x)} \neq 0#, whereas the domain of #\green{s(x)}# contains all numbers #x#. We can only say that #f(x) = \green{s(x)}# if we add the restriction that #\orange{q(x)} \neq 0#.

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} \begin{array}{l} \orange{q(x)} \enclose{longdiv}{\blue{p(x)}} \end{array}#	\[ \require{enclose} \begin{array}{l} \orange{x+1} \enclose{longdiv}{\blue{2x^2+5x+5}} \end{array}\]
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} \begin{array}{l} \phantom{x+1\hspace{0.5em}} \green{2x+3} \\ \orange{x+1} \enclose{longdiv}{\blue{2x^2+5x+5}} \\ \phantom{x+1\hspace{0.5em}} \underline{2x^2+2x} \;\underline{\;} \\ \phantom{x+1\hspace{1.8em}2x^2} {\purple{3x+5}}\\ \phantom{x+1\hspace{1.8em}2x^2} \underline{3x+3}\;\underline{\;}\\ \phantom{x+1\hspace{4.1em}2x^2}\purple{2} \end{array}#
Step 3	It now follows that \[ \begin{array}{rcl} f(x) &=& \dfrac{\blue{p(x)}}{\orange{q(x)}} \\ &=&\green{s(x)}+\dfrac{\purple{r(x)}}{\orange{q(x)}}\end{array} \]	\[\begin{array}{rcl} f(x) &=&\dfrac{\blue{2x^2+5x+5}}{\orange{x+1}} \\ &=&\green{2x+3} + \dfrac{\purple{2}}{\orange{x+1}} \end{array}\]

Use long division to write the function

\[f(x)=\frac{4x+4}{4x+5}\]

in the form

\[f(x)=s+\frac{r}{ax+b}\]

where #a, b,r# and #s# are constants.

#f(x) = \;##{1} + \dfrac{{-1}}{{4x+5}}#

Step 1

We first compose the long division: \[\require{enclose}
\begin{array}{l}
{4x+5} \enclose{longdiv}{{4x+4}}
\end{array}\]

Step 2

Long division gives

\[ \require{enclose}
\begin{array}{l}
\phantom{4x+5\hspace{0.5em}} {1} \\
{4x+5} \enclose{longdiv}{{4x+4\phantom{\;}}} \\
\phantom{4x+5\hspace{0.5em}} \underline{4 x+5} \;\underline{\phantom{\;}} \\
\phantom{4x+5\hspace{0.5em}ax\:\:}{-1}
\end{array}\]

Step 3

According to the theory, it now follows that

\[\begin{array}{rcl} f(x) &=&\dfrac{{4x+4}}{{4x+5}} \\
&=&{1} + \dfrac{{-1}}{{4x+5}}
\end{array}\]

New example