\chapter{Polynomial and rational functions}
\section{The remainder and factor theorems}
In the previous unit, we saw that synthetic division is an efficient way to divide a given
polynomial by a first degree binomial such as \(x-3\) or \(x+2\). We will now
focus on the remainder of synthetic division. The remainder will be a constant (without variables).
We will see how this constant is related to the polynomial we divided.
\subsection{The remainder theorem}
\begin{example}
Let's work the first example of the previous unit again,
the division of \(f(x)=3x^3-2x^2+x-4\) by \(x-2\):
| \(2\) |
\(3\) | \(-2\) | \(1\) | \(-4\) |
| | | \(6\) | \(8\) | \(18\) |
| | \(3\) | \(4\) | \(9\) | \(\mathbf{14}\) |
So the remainder is \(14\).
Now compute \(f(2)\):
\[
\begin{array}{rcl}
f(2) &=& 3(2)^3-2(2)^2+2-4\\[1ex]
&=& 24-8-2\\[1ex]
&=& 14
\end{array}
\]
So \(f(2)\) is the same as the remainder when we divide \(f(x)\) by \(x-2\).
Now we divide \(f(x)\) by \(x+3\):
| \(-3\) |
\(3\) | \(-2\) | \(1\) | \(-4\) |
| | | \(-9\) | \(33\) | \(-102\) |
| | \(3\) | \(-11\) | \(34\) | \(\mathbf{-106}\) |
The remainder is \(-106\). Compute \(f(-3)\):
\[
\begin{array}{rcl}
f(-3)&=& -3(3)^3-2(-3)^2-3-4\\[1ex]
&=& -81-18-3-4\\[1ex]
&=& -106
\end{array}
\]
Again, \(f(-3)\) is the same as the remainder when we divide \(f(x)\) by \(x+3\).
\end{example}
The previous examples illustrate something that is always true: the \textit{Remainder Theorem}.
| Remainder Theorem |
|---|
| If the polynomial \(f(x)\) is divided by \(x-c\),
the remainder is \(f(c)\). |
It is easy to understand why the Remainder Theorem is true: remember that we can write the
result of the synthetic division (see the beginning of Unit 2.5) as:
\eqlabel{Pdiv} \begin{equation} f(x)=q(x)(x-c) +r. \end{equation}
Then, substituting \(x=c\) in the above equation, we get
\[f(c)=q(c)(c-c)+r=0+r=r.\]
Think of the Remainder Theorem as an alternative way to find function values: if you are given
a polynomial \(f(x)\) and want to find \(f(c)\), you could put \(x=c\) in the formula for \(f(x)\)
(as we did at the beginning of Chapter 1), or you could divide \(f(x)\) by \(x-c\) and look at the
remainder. The first method makes no use of the Remainder Theorem, but the second does.
So to work problems in this section correctly, it is important to look at the directions carefully,
so that you know which method to use. And in cases where you have a choice of methods, it could be
that one method is easier and faster than the other.
\begin{example}
Let \(f(x)=x^4+10x^3+8x^2-8x+7\). We want to use the Remainder Theorem to find the function value
\(f(-9)\). We could put \(x=-9\) in the given formula for \(f(x)\). But then
we would not be using the Remainder Theorem.
Instead, we divide by \((x+9)\) with synthetic division and look
at the remainder. The Remainder Theorem then tells us that the remainder is the same as \(f(-9)\).
| \(-9\) |
\(1\) | \(10\) | \(8\) | \(-8\) | \(7\) |
| | | \(-9\) | \(-9\) | \(9\) | \(-9\) |
| | \(1\) | \(1\) | \(-1\) | \(1\) | \( \mathbf{-2}\) |
So the remainder is \(-2\), and \(f(-9)=-2\).
\end{example}
Note that in the previous example the synthetic division was simple, but substituting \(x=-9\)
in \(f(x)\) would involve finding the \(4\)-th power of \(-9\).
In a similar way, if we are given a polynomial \(f(x)\) and are asked to find the remainder
when we divide it by \(x-c\), we could do synthetic division and look at the remainder.
But this would not be using the Remainder Theorem. Instead, if we plug in \(x=c\) in the
formula for \(f(x)\), then the Remainder Theorem tells us that \(f(c)\) is the same as the remainder.
\begin{example}
Let \(f(x)=7x^{15}-2x^8-2x^3-1\). We want to use the Remainder Theorem to find the
remainder when \(f(x)\) is divided by \(x+1\). If we do synthetic division, we would not be
using the Remainder Theorem. Besides, the polynomial has degree \(15\) with a lot of missing
powers, so we would need a large synthetic division table. Instead, we plug in \(x=-1\):
\[
\begin{array}{rcl}
f(-1)&=& 7(-1)^{15} -2(-1)^8-2(-1)^3-1\\[1ex]
&=& -7-2+2-1\\[1ex]
&=& -8
\end{array}
\]
So \(f(-1)=-8\), and the Remainder Theorem says that if we did the synthetic division, we would
find the remainder to be \(-8\).
\end{example}
Having two different methods to find an answer also gives us a way to check that our answer is
correct. So we can also use the Remainder Theorem as a method of checking our answer, as in the
next example.
\begin{example}
Let \(f(x)=4x^3-9x^2-6x-7\). We find \(f(3)\), then use the Remainder Theorem to check the answer.
Substituting \(x=3\) we find
\[
\begin{array}{rcl}
f(3)&=& 4(3)^3 -9(3)^2-6(3)-7\\[1ex]
&=& 108-81-18-7\\[1ex]
&=& 2
\end{array}
\]
Now we check the answer by doing synthetic division:
| \(3\) |
\(4\) | \(-9\) | \(-6\) | \(-7\) |
| | | \(12\) | \(9\) | \(9\) |
| | \(4\) | \(3\) | \(3\) | \( \color{red}{\mathbf{2}}\) |
The remainder is the same as the function value \(f(3)\), as the remainder theorem says.
\end{example}
\subsection{The factor theorem}
As we mentioned earlier in this unit (see Equation (\eqref{Pdiv})), when we divide
a polynomial \(f(x)\) by \(x-c\), we
can write the answer as
\[f(x)=q(x)(x-c)+r.\]
\begin{example}
If we divide \(f(x)=3x^2-4x+5\) by \(x-3\) we find
| \(3\) |
\(3\) | \(-4\) | \(5\) |
| | | \(9\) | \(15\) |
| | \(3\) | \(5\) |
\( \mathbf{20}\) |
so the quotient is \(q(x)=3x+5\) and the remainder \(r=20\), and we can write
\[3x^2-4x+5=(3x+5)(x-3)+20\]
\end{example}
\begin{example}
Now look at a second example: We divide \(3x^2-4x-15\) by \(x-3\)
| \(3\) |
\(3\) | \(-4\) | \(-15\) |
| | | \(9\) | \(15\) |
| | \(3\) | \(5\) |
\( \color{red}{\mathbf{0}}\) |
This time the remainder is zero. So we write the answer as
\[3x^2-4x-15=(3x+5)(x-3)\]
But this means that the polynomial \(3x^2-4x-15\)
has been factored, and \((x-3)\) is one of the factors.
\end{example}
The previous example illustrates an important consequence of the Remainder Theorem:
because the remainder (when we divide \(f(x)\) by \(x-c\)) is \(f(c)\), we can conclude that
if \(f(c)=0\) then there is no remainder, and so \(x-c\) will be a factor of \(f(x)\). This is
in fact the statement of the Factor Theorem, that we spell out below:
| Factor Theorem |
|---|
| \(x-c\) is a factor of \(f(x)\) precisely when \(f(c)=0\). |
This is not really new: after all, when
we look for the \(x\)-intercepts of a polynomial \(f(x)\), we set \(f(x)=0\), and try to solve the
equation. We know that if we are able to factor it, then each factor can be set to zero to
find the \(x\)-intercepts.
So knowing that \(x-c\) is a factor of \(f(x)\) is the same as
knowing that \((c,0)\) is an \(x\)-intercept of \(f(x)\), and we have been using the
Factor Theorem all along whenever we solved \(f(x)=0\) by factoring.
But the new (and useful) information we get in this unit is that to know if \((x-c)\) is a factor, we
can do synthetic division and see if we get a zero remainder.
\begin{example}
Let \(f(x)=x^4 - 48x^2 + 8x + 10\). We would like to find out if \((-7,0)\) is an \(x\)-intercept
of \(f(x)\).
We could plug in \(x=-7\) and check. But we use the Factor Theorem instead and divide by \(x+7\)
using synthetic division (remember that this is sometimes easier and faster than plugging in):
| \(-7\) |
\(1\) | \(0\) | \(-48\) |
\(8\) | \(10\) |
| | | \(-7\) | \(49\) | \(-7\) | \(-7\) |
| | \(1\) | \(-7\) |
\(1\) | \(1\) |
\(\mathbf{3}\) |
We found that the remainder is not zero. So we conclude that \(x+7\) is not a factor, and
\((-7,0)\) is not an \(x\)-intercept.
\end{example}
\begin{example}
Let \(f(x)=x^4 - 7x^3 - 9x^2 + 9x - 8\). We want to know if \((8,0)\) is an \(x\)-intercept.
We divide by \(x-8\) using synthetic division:
| \(8\) |
\(1\) | \(-7\) | \(-9\) |
\(9\) | \(-8\) |
| | | \(8\) | \(8\) | \(-8\) | \(8\) |
| | \(1\) | \(1\) |
\(-1\) | \(1\) |
\( \mathbf{0}\) |
This time we found that the remainder is zero. So we conclude that \(x-8\) is a factor, and
\((8,0)\) is an \(x\)-intercept.
\end{example}
There are cases when plugging in a number is faster than doing synthetic division, for example
it is quite easy to plug in \(x=1\) or \(x=-1\), and if the degree of the polynomial is high,
the synthetic division requires a lot of entries.
\begin{example}
Let \(f(x)=x^{12} - 2x^{11} + x^{10} + x^2 - 2x + 1\). We would like to know if \(x-1\) is a factor
of \(f(x)\). We could do synthetic division for this \(12\)-th degree polynomial
and see if the remainder is zero. But the Factor
Theorem gives us a better way: plug in \(x=1\) and see if we get zero:
\[
\begin{array}{rcl}
f(1)&=& (1)^{12} -2(1)^{11}+(1)^{10} +(1)^2-2(1)+1\\[1ex]
&=& 1-2+1+1-2+1\\[1ex]
&=& 0.
\end{array}
\]
Since \(f(1)=0\), we conclude that \(x-1\) is a factor of \(f(x)\).
Next we want to know if \(x+1\) is a factor. Again, instead of doing synthetic division, we
plug in \(x=-1\):
\[
\begin{array}{rcl}
f(-1)&=& (-1)^{12} -2(-1)^{11}+(-1)^{10} +(-1)^2-2(-1)+1\\[1ex]
&=& 1+2+1+1+2+1\\[1ex]
&=& 8.
\end{array}
\]
This time \(f(-1)\) is not zero, and so \(x+1\) is not a factor.
\end{example}
Problems
\problem
Let \(f(x)=2x^4 - 20x^3 - 23x^2 + 7x + 11\). Use the Remainder Theorem to find \(f(11)\).
\begin{sol}
We use synthetic division to divide \(f(x)\) by \(x-11\):
| \(11\) |
\(2\) | \(-20\) | \(-23\) | \(7\) | \(11\) |
| | | \(22\) | \(22\) | \(-11\) | \(-44\) |
| | \(2\) | \(2\) | \(-1\) | \(-4\) | \(\mathbf{-33}\) |
The remainder is \(-33\). So the Remainder Theorem tells us that \(f(11)=-33\).
\end{sol} \mproblem
Let \(f(x)=3x^4+35x^3-9x^2+37x+14\). Use the Remainder Theorem to find \(f(-12)\).
\problem
Let \(f(x)=12x^9-3x^2-8\). Use the Remainder Theorem to find the remainder
when \(f(x)\) is divided by \(x-2\).
\begin{sol}
We find \(f(2)\):
\[
\begin{array}{rcl}
f(2)&=& 12(2)^9 -3(2)^2-8\\[1ex]
&=& 6144-12-8\\[1ex]
&=& 6124.
\end{array}
\]
So the remainder when dividing by \(x-2\) will be 6124.
\end{sol} \mproblem
Let \(f(x)=8x^{11}-9x^4+7x-3\). Use the Remainder Theorem to find the remainder when \(f(x)\)
is divided by \(x+1\).
\problem
Let \(f(x)=7x^4-3x^3+x^2-9\). Find \(f(-2)\), then use the Remainder Theorem to check your answer.
\begin{sol}
\[
\begin{array}{rcl}
f(-2)&=& 7(-2)^4-3(-2)^3+(-2)^2-9\\[1ex]
&=& 112+24+4-9\\[1ex]
&=& 131
\end{array}
\]
Now we use synthetic division. If our answer is right, the remainder must be 131.
| \(-2\) |
\(7\) | \(-3\) | \(1\) |
\(0\) | \(-9\) |
| | | \(-14\) | \(34\) | \(-70\) | \(140\) |
| | \(7\) | \(-17\) |
\(35\) | \(-70\) |
\( \mathbf{131}\) |
\end{sol} \mproblem
Let \(f(x)=5x^4-8x^3+2x^2-1\). Find \(f(3)\), then use the Remainder Theorem to check the answer.
\problem
Let \(f(x)=x^4 + 2x^3 - 25x^2 - 2x + 24\)
\begin{enumerate}
\item Use the Factor Theorem to decide if \((-6,0)\) is an \(x\)-intercept for \(f(x)\).
\item Repeat part a. using \((-2,0)\).
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item We divide \(f(x)\) by \(x+6\) using synthetic division:
| \(-6\) |
\(1\) | \(2\) | \(-25\) |
\(-2\) | \(24\) |
| | | \(-6\) | \(24\) | \(6\) | \(-24\) |
| | \(1\) | \(-4\) |
\(-1\) | \(4\) |
\(\mathbf{0}\) |
The remainder is zero, and we conclude \((-6,0)\) is an \(x\)-intercept.
\item We do the same for \((-2,0)\). We divide \(f(x)\) by \(x+2\) using synthetic division:
| \(-6\) |
\(1\) | \(2\) | \(-25\) |
\(-2\) | \(24\) |
| | | \(-2\) | \(0\) | \(50\) | \(-96\) |
| | \(1\) | \(0\) |
\(-25\) | \(48\) |
\(\mathbf{-72}\) |
This time the remainder is not zero, so we conclude that \((-2,0)\) is not an \(x\)-intercept.
\end{enumerate}
\end{sol} \mproblem
Let \(f(x)=x^4 + 6x^3 - 12x^2 - 38x - 21\).
\begin{enumerate}
\item Use the Factor Theorem to decide if \((-3,0)\) is an \(x\)-intercept for \(f(x)\).
\item Repeat part a. using \((-7,0)\).
\end{enumerate}
\problem
Let \(f(x)=x^{13} + 2x^{12} + x^{11} - 2x^2 - 4x - 2\).
\begin{enumerate}
\item Use the Factor Theorem to decide if \(x-1\) is a factor of \(f(x)\).
\item Repeat part (a) for \(x+1\).
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
The directions say that we need to use the Factor Theorem.
If we divide by \(x-1\) with synthetic division, we
can find out if the remainder is zero, but we would not be using the Factor Theorem.
Besides, it would be a long computation with a \(13\)-th degree polynomial.
Instead, we compute \(f(1)\):
\[
\begin{array}{rcl}
f(1)&=& (1)^{13} + 2(1)^{12} +(1)^{11} - 2(1)^2-4(1) -2 \\[1ex]
&=& 1+2+1-2-4-2\\[1ex]
&=& -4.
\end{array}
\]
We found that \(f(1)\) is not zero.
So \(x-1\) is not a factor.
\item
We compute \(f(-1)\):
\[
\begin{array}{rcl}
f(-1)&=& (-1)^{13} + 2(-1)^{12} +(-1)^{11} - 2(-1)^2-4(-1) -2 \\[1ex]
&=& -1+2-1-2+4-2\\[1ex]
&=& 0.
\end{array}
\]
Since \(f(-1)=0\), \(x+1\) is a factor of \(f(x)\).
\end{enumerate}
\end{sol} \mproblem
Let \(f(x)=x^{11} + 2x^{10} -x^9 + 6x^2 -2x -6\).
\begin{enumerate}
\item Use the Factor Theorem to decide if \(x-1\) is a factor of \(f(x)\).
\item Repeat part (a) for \(x+1\).
\end{enumerate}