\chapter{Polynomial and rational functions}
\section{Division of polynomials}
In this unit we will study how to divide polynomials. To understand what this really means,
let's review what division of whole numbers means.
Suppose we want to divide 38 by 5. Then
of course we cannot do it exactly. The largest multiple of 5
that is no larger than 38 is 7, and that is called the quotient. This means that there will be
a remainder of 3, because \(7\times 5=35\), and we can write \(38=7\times 5 +3\).
It also means that we can write the result of the division as
\[\frac{38}{5}=7+\frac{3}{5}.\]
The remainder must be smaller than what we are dividing by (5 in this case),
because otherwise we could have increased the quotient by 1.
Division of polynomials works the same way, except that to find the right remainder,
instead of the size of the numbers we look at the degrees of the polynomials.
So if we are given two polynomials \(f(x)\) and \(g(x)\), to divide \(f(x)\) by \(g(x)\)
means finding a quotient polynomial \(q(x)\) and a remainder polynomial \(r(x)\) that
allow us to write
\[\frac{f(x)}{g(x)} = q(x) +\frac{r(x)}{g(x)}\]
where the degree of \(r(x)\) must be smaller than the degree of \(g(x)\).
Multiplying the preceding equation by \(g(x)\), we can re-write it as
\[f(x)=q(x)g(x) +r(x)\]
and this second form of the division will be especially useful in the next unit.
We now look at two methods to find the quotient and the remainder for polynomials.
\subsection{Synthetic division}
\textit{Synthetic division} is an efficient way to divide a given polynomial by a first degree binomial
of the form \(x-c\) or \(x+c\), such as \(x-3\) or \(x+2\). Since the degree of such a binomial
is 1, the remainder will have to have degree 0. That means that the remainder will be a constant.
Look at the following example.
\begin{example}
Let \(f(x)=3x^3-2x^2+x-4\). We divide \(f(x)\) by \(x-2\) using synthetic division. Begin by drawing a
box as shown, and placing the coefficients of \(f(x)\) at the top, right of the vertical line.
| |
\(3\) | \(-2\) | \(1\) | \(-4\) |
| | | | | |
| | | | | |
Then, if we are dividing by \(x-c\), we place the number \(c\) at the top, left of the vertical line.
In our case we are dividing by \(x-2\) so we place \(2\). Notice the changed sign. If we were dividing
by \(x+c\) we would place \(-c\).
| \(2\) |
\(3\) | \(-2\) | \(1\) | \(-4\) |
| | | | | |
| | | | | |
The next step is simply to carry down the first entry right of the vertical line (\(3\) in our case)
to the bottom:
| \(2\) |
\(3\) | \(-2\) | \(1\) | \(-4\) |
| | \(\downarrow\) | | | |
| | \(3\) | | | |
Now multiply the number at the bottom by the number to the left of the vertical line (both shown in
bold) and place the result under the second number at the top inside the box:
| \( \mathbf{2}\) |
\(3\) | \(-2\) | \(1\) | \(-4\) |
| | | \(\color{blue}{6}\) | | |
| | \(\mathbf{3}\) | | | |
Next we add the number we just placed to the one above it (both shown in bold),
and place the result at the bottom:
| \(2\) |
\(3\) | \(\mathbf{-2}\) | \(1\) | \(-4\) |
| | | \(\mathbf{6}\) | | |
| | \(3\) | \(\mathbf{4}\) | | |
Now we repeat the process using the new number at the bottom: we multiply \(4\) by \(2\), place the
result under the next number inside the box, add with the one above it, and write the answer at
the bottom:
| \(2\) |
\(3\) | \(-2\) | \(1\) | \(-4\) |
| | | \(6\) | \(8\) | |
| | \(3\) | \(4\) | \(9\) | |
We do it one more time using the new number at the bottom (\(9\times 2=18\)),
and that completes the synthetic division:
| \(2\) |
\(3\) | \(-2\) | \(1\) | \(-4\) |
| | | \(6\) | \(8\) | \(18\) |
| | \(3\) | \(4\) | \(9\) | \( \mathbf{14}\) |
The number at the right of the bottom row (shown in bold) is the remainder.
The other numbers on the bottom
row are the coefficients of the quotient polynomial \(q(x)\),
that will always be of degree one less than \(f(x)\). So in this example, since \(f(x)\)
has degree \(3\), the quotient has degree \(2\): \(q(x)=3x^2+4x+9\).
The remainder is \(r=14\), and we can write the answer
of the polynomial division as
\[\frac{3x^3-2x^2+x-4}{x-2}=3x^2+4x+9+\frac{14}{x-2}.\]
\end{example}
If some powers of \(x\) are missing in \(f(x)\), we must make sure to insert a \(0\) in their
place when we set up the synthetic division, as shown in the next example.
\begin{example}
We divide \(f(x)=4x^4+12x^3+5x+9\) by \(x+3\).
Because the power \(x^2\) is missing, we insert a
\(0\) in its place:
| \(-3\) |
\(4\) | \(12\) | \(0\) | \(5\) | \(9\) |
| | | | | |
|
| | | | | | |
Then we continue filling the table as before:
| \(-3\) |
\(4\) | \(12\) | \(0\) | \(5\) | \(9\) |
| | | \(-12\) | \(0\) | \(0\) |
\(-15\) |
| | \(4\) | \(0\) | \(0\) | \(5\) | \(-6\) |
So the remainder is \(r=-6\) and the quotient is \(q(x)=4x^3+0x^2+0x+5=4x^3+5\), and we can write the
division as
\[\frac{4x^4+12x^3+5x+9}{x+3}=4x^3+5 -\frac{6}{x+3}.\]
\end{example}
\subsection{Long division}
If we want to divide by a polynomial of degree \(2\) or more, then we need to use \textit{long division}.
We show how it works by an example.
\begin{example}
Suppose we want to divide \(f(x)=6x^4 + 8x^3 + 3x^2 - 4\) by \(g(x)=3x^2-2x+1\).
We begin by drawing a box as shown, with the polynomial \(f(x)\) written inside, and \(g(x)\) on the
left side. This time we also need to keep the variable \(x\), and as before we need to insert
\(0\)'s whenever a power is missing. Leave some space at the top, because that is where we
will write the quotient.
| |
| |
|
|
|
|
|
| \(3x^2\) | \(-2x\) | \(+1\) |
\(6x^4\) |
\(+8x^3\) | \(+3x^2\) | \(+0x\) |
\(-4\) |
The first step is to divide the leading term of \(f(x)\) by the leading term of \(g(x)\)
(both shown in bold in the next table), and place the result at the top. This will
be the leading term of the quotient.
\[\mathbf{\frac{6x^4}{3x^2}}=2x^2.\]
| | | |
\(\mathbf{2x^2}\) | | | | |
| \(\mathbf{3x^2}\) | \(-2x\) | \(+1\) |
\(6x^4\) |
\(+8x^3\) | \(+3x^2\) | \(+0x\) |
\(-4\) |
The next step is to multiply the term that we just placed at the top by every term of \(g(x)\)
(on the left of the box), and place the result under the terms of \(f(x)\) inside the box,
making sure that all the powers of \(x\) are aligned correctly (meaning that powers of \(x^4\)
should be under powers of \(x^4\), powers of \(x^3\) under powers of \(x^3\), and so on.
| | | |
\(\mathbf{2x^2}\) | | | | |
| \(\mathbf{3x^2}\) |
\(\mathbf{-2x}\) |
\(\mathbf{+1}\) |
\(6x^4\) |
\(+8x^3\) | \(+3x^2\) | \(+0x\) |
\(-4\) |
| | | |
\(6x^4\) |
\(-4x^3\) |
\(+2x^2\) | | |
Now we change the sign of the terms we just placed, draw a horizontal line,
and add with the terms above, writing the
result under the line. Make sure to also bring down under the line the remaining terms of \(f(x)\).
| | | |
\(2x^2\) | | | | |
| \(3x^2\) | \(-2x\) | \(+1\) |
\(6x^4\) |
\(+8x^3\) | \(+3x^2\) | \(+0x\) |
\(-4\) |
| change | sign: | |
\(-6x^4\) |
\(+4x^3\) | \(-2x^2\) |
| |
| | | |
| \(12x^3\) | \(+x^2\) | \(+0x\) | \(-4\) |
The polynomial at the bottom is of degree 3, one lower than \(f(x)\), but still higher than \(g(x)\).
We will keep repeating
the process until the polynomial at the bottom has a degree lower than \(g(x)\).
So we divide the leading term of the polynomial at the bottom
by the leading term of \(g(x)\) (both shown in bold) and the result (placed at the top)
is the second term for the quotient:
\[ \mathbf{\frac{12x^3}{3x^2}}=4x\]
| | | |
\(2x^2 \) | \(+4x\) |
| | |
| \( \mathbf{3x^2}\) | \(-2x\) | \(+1\) |
\(6x^4\) |
\(+8x^3\) | \(+3x^2\) | \(+0x\) |
\(-4\) |
| | | |
\(-6x^4\) |
\(+4x^3\) | \(-2x^2\) |
| |
| | | |
| \(\mathbf{12x^3}\) | \(+x^2\) | \(+0x\) | \(-4\) |
We multiply the new term just placed at the top by each term of \(g(x)\) (on the left), and
place the result at the bottom, again making sure to align like powers of \(x\):
| | | |
\(2x^2 \) | \(\mathbf{+4x}\) |
| | |
| \(\mathbf{3x^2}\) |
\(\mathbf{-2x}\) |
\(\mathbf{+1}\) |
\(6x^4\) |
\(+8x^3\) | \(+3x^2\) | \(+0x\) |
\(-4\) |
| | | |
\(-6x^4\) |
\(+4x^3\) | \(-2x^2\) |
| |
| | | |
| \( 12x^3\) | \(+x^2\) | \(+0x\) | \(-4\) |
| | | |
| \(12x^3\) |
\(-8x^2\) |
\(+4x\) | |
We change all signs in the polynomial at the bottom, draw a line, add with the previous line and
write the result under the line:
| | | |
\(2x^2 \) | \(+4x\) | | | |
| \(3x^2\) | \(-2x\) | \(+1\) |
\(6x^4\) |
\(+8x^3\) | \(+3x^2\) | \(+0x\) |
\(-4\) |
| | | |
\(-6x^4\) |
\(+4x^3\) | \(-2x^2\) |
| |
| | | |
| \( 12x^3\) | \(+x^2\) | \(+0x\) | \(-4\) |
| | change | sign: |
|
\(-12x^3\) |
\( +8x^2\) |
\( -4x\) |
|
| | | |
| |
\(9x^2\) |
\(-4x\) |
\(-4\) |
The polynomial at the bottom is now of degree 2, the same as \(g(x)\). So we can do one more step.
(We stop when the degree gets smaller than the degree of \(g(x)\)).
Divide the leading terms and add the result to the quotient at the top:
\[\mathbf{\frac{9x^2}{3x^2}}=3\]
| | | |
\(2x^2\) | \(+4x\) |
\(\color{blue}{\mathbf +3}\) |
| |
| \(\mathbf{3x^2}\) | \(-2x\) | \(+1\) |
\(6x^4\) |
\(+8x^3\) | \(+3x^2\) | \(+0x\) |
\(-4\) |
| | | |
\(-6x^4\) |
\(+4x^3\) | \(-2x^2\) |
| |
| | | |
| \( 12x^3\) | \(+x^2\) | \(+0x\) | \(-4\) |
| | | |
| \(-12x^3\) |
\( +8x^2\) |
\( -4x\) |
|
| | | |
| |
\(\mathbf{9x^2}\) |
\(-4x\) |
\(-4\) |
Now multiply by each term of \(g(x)\), place at the bottom, change the sign, and add with the
previous line:
| | | |
\(2x^2\) | \(+4x\) |
\(+3\) |
| |
| \( 3x^2\) |
\(-2x\) |
\(+1\) |
\(6x^4\) |
\(+8x^3\) | \(+3x^2\) | \(+0x\) |
\(-4\) |
| | | |
\(-6x^4\) |
\(+4x^3\) | \(-2x^2\) |
| |
| | | |
| \( 12x^3\) | \(+x^2\) | \(+0x\) | \(-4\) |
| | | |
| \(-12x^3\) |
\( +8x^2\) |
\( -4x\) |
|
| | | |
| |
\(9x^2\) |
\(-4x\) |
\(-4\) |
| | | |
| |
\( -9x^2\) |
\(+6x\) |
\(-3\) |
| | | |
| |
|
\(2x\) |
\( -7\) |
The polynomial at the bottom now has degree 1, lower than \(g(x)\), so we stop. The quotient
of the long division is at the top, and the remainder at the bottom. So we can write
\[\frac{6x^4 + 8x^3 + 3x^2 - 4}{3x^2-2x+1}=2x^2+4x+3+\frac{2x-7}{3x^2-2x+1}.\]
\end{example}
Problems
\problem
Use synthetic division to perform the polynomial divisions.
\begin{enumerate}
\item
\(\displaystyle \frac{2x^3-x^2+7}{x+2}\)
\item
\(\displaystyle \frac{x^4-2x^3+3x^2-4x+5}{x+1}\)
\item
\(\displaystyle \frac{x^5-3}{x-3}\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
\(\displaystyle \frac{2x^3-x^2+7}{x+2}\).
The power \(x\) is missing, so we need to insert a \(0\) in its place:
| \(-2\) |
\(2\) | \(-1\) | \(0\) |
\(7\) |
| | | \(-4\) | \(10\) | \(-20\) |
| | \(2\) | \(-5\) |
\(10\) | \( -13\) |
So the answer is
\[\frac{2x^3-x^2+7}{x+2}=2x^2-5x+10 -\frac{13}{x+2}.\]
\item
\(\displaystyle \frac{x^4-2x^3+3x^2-4x+5}{x+1}\).
The synthetic division table is:
| \(-1\) |
\(1\) | \(-2\) | \(3\) |
\(-4\) | \(5\) |
| | | \(-1\) | \(3\) | \(-6\) |
\(10\) |
| | \(1\) | \(-3\) | \(6\) | \(-10\) | \(15\) |
and the answer is
\[\frac{x^4-2x^3+3x^2-4x+5}{x+1}=x^3-3x^2+6x-10 +\frac{15}{x+1}.\]
\item
\(\displaystyle \frac{x^5-3}{x-3}\). There are four missing powers: \(x^4\), \(x^3\), \(x^2\), \(x\),
so we need to insert
\(0\)'s in their place:
| \(3\) |
\(1\) | \(0\) | \(0\) |
\(0\) | \(0\) | \(-3\) |
| | | \(3\) | \(9\) | \(27\) |
\(81\) |
\(243\) |
| | \(1\) | \(3\) | \(9\) | \(27\) | \(81\) | \(240\) |
The answer is
\[\frac{x^5-3}{x-3}=x^4+3x^3+9x^2+27x+81 +\frac{240}{x-3}.\]
\end{enumerate}
\end{sol} \mproblem
Use synthetic division to perform the polynomial divisions.
\begin{enumerate}
\item
\(\displaystyle \frac{5x^4-x^2+x+1}{x+3}\)
\item
\(\displaystyle \frac{3x^3+4}{x+4}\)
\item
\(\displaystyle \frac{x^5-x^4+x^3-x^2+x-1}{x-2}\)
\end{enumerate}
\problem
Use long division to perform the polynomial divisions.
\begin{enumerate}
\item
\(\displaystyle \frac{9x^3 - 8x + 13}{3x^2-4x+2}\)
\item
\(\displaystyle \frac{6x^4 + x^3 - 3x^2 - 2}{3x^2+2x+1}\)
\item
\(\displaystyle \frac{5 x^5 - 2}{5 x^4 - 10 x^3 + 5}\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
\(\displaystyle \frac{9x^3 - 8x + 13}{3x^2-4x+2}\).
| | | |
\(3x\) | \(+4\) |
|
| \( 3x^2\) |
\(-4x\) |
\(+2\) |
\(9x^3\) |
\(+0x^2\) | \(-8x\) | \(+13\) |
| | | |
\(-9x^3\) |
\(+12x^2\) | \(-6x\) |
|
| | | |
| \( 12x^2\) | \(-14x\) | \(+13\) |
| | | |
| \(-12x^2\) |
\( +16x\) |
\( -8\) |
| | | |
| |
\(2x\) |
\(+5\) |
\[ \frac{9x^3 - 8x + 13}{3x^2-4x+2}=3x+4 +\frac{2x+5}{3x^2-4x+2}.\]
\item
\(\displaystyle \frac{6x^4 + x^3 - 3x^2 - 2}{3x^2+2x+1}\).
| | | |
\(2x^2\) | \(-x\) |
\(-1\) |
| |
| \( 3x^2\) |
\(+2x\) |
\(+1\) |
\(6x^4\) |
\(+x^3\) | \(-3x^2\) | \(+0x\) |
\(-2\) |
| | | |
\(-6x^4\) |
\(-4x^3\) | \(-2x^2\) |
| |
| | | |
| \( -3x^3\) | \(-5x^2\) | \(+0x\) | \(-2\) |
| | | |
| \(3x^3\) |
\( +2x^2\) |
\( +x\) |
|
| | | |
| |
\(-3x^2\) |
\(+x\) |
\(-2\) |
| | | |
| |
\( 3x^2\) |
\(+2x\) |
\(+1\) |
| | | |
| |
|
\(3x\) |
\( -1\) |
\[\frac{6x^4 + x^3 - 3x^2 - 2}{3x^2+2x+1}=2x^2-x-1+\frac{3x-1}{3x^2+2x-1}\]
\item
\(\displaystyle \frac{5 x^5 - 2}{5 x^4 - 10 x^3 + 5}\).
| |
|
|
| |
\(x\) | \(+2\) |
|
| |
|
| \( 5x^4\) |
\(-10x^3\) |
\(+0x^2\) |
\(+0x\) |
\(+5\) |
\(5x^5\) |
\(+0x^4\) | \(+0x^3\) | \(+0x^2\) |
\(+0x\) |
\(-2\) |
| |
|
| | |
\(-5x^5\) |
\(+10x^4\) | \(-0x^3\) |
\(-0x^2\) | \(-5x\) |
|
| |
|
| | |
| \( 10x^4\) | \(0x^3\) | \(+0x^2\) | \(-5x\) |
\(-2\) |
| |
|
| | |
| \(-10x^4\) |
\( +20x^3\) |
\( -0x^2\) |
\(-0x\) |
\(-10\) |
| |
|
| | |
| |
\(20x^3\) |
\(+0x^2\) |
\(-5x\) |
\(-12\) |
\[\frac{5 x^5 -2}{5 x^4 - 10 x^3 + 5}=x+2+\frac{20x^3-5x-12}{5x^4-10x^3+5}.\]
\end{enumerate}
\end{sol} \mproblem
Use long division to perform the polynomial divisions.
\begin{enumerate}
\item
\(\displaystyle \frac{8x^3 + 20x^2 - 15x - 36}{4x^2-7}\)
\item
\(\displaystyle \frac{4x^4 - 11x^2 + 16x - 8}{2x^2-x+1}\)
\item
\(\displaystyle \frac{ 3x^5 + 27x^4 + 9x^3 - 24x^2 + 7x + 5}{3x^3-3x+2}\)
\end{enumerate}