\chapter{Polynomial and rational functions}
\section{The rational zero theorem}
\subsection{Zeros of polynomial functions}
Remember from Chapter 1 that a zero (or root) of a polynomial function \(f(x)\) is a solution of the
equation \(f(x)=0\).
Considering the graph of \(f(x)\), a number \(c\) is a zero precisely when
\((c,0)\) is an \(x\)-intercept.
We know from Unit 2.4 that the graph of a polynomial function of degree \(n\) has at most
\(n\) \(x\)-intercepts. So we conclude:
| A polynomial function of degree \(n\) has at most \(n\) zeros. |
For example, a quadratic function (studied in Unit 2.1 and Unit 2.2), has degree \(2\).
So it has at most two zeros, and we know two main methods to find them: either by factoring, or
(if it does not factor) with the quadratic formula.
But there is an important difference between the two methods: if we cannot factor a
quadratic polynomial whose coefficients are whole numbers, the zeros that we get from
the quadratic formula will involve some square roots.
For higher degree polynomials, it can be quite difficult to find the zeros.
Even for a simple looking third degree polynomial such as \(x^3+x+1\),
it is not at all easy to find the zeros.
In this unit we will discuss the problem of finding the zeros of polynomial functions of degree
larger than 2.
But to do this we first need to review the different kinds of numbers.
The \textit{rational numbers} are all the fractions that have whole numbers in both numerator and denominator.
So for example
\[\frac{1}{2}, \ \ \ 1, \ \ \ -\frac{3}{5}, \ \ \ 0\]
are rational numbers, while
\[\sqrt{2}, \ \ \ \pi,\ \ \ \frac{\pi}{4}\]
are real numbers, but are not rational (also called \textit{irrational numbers}).
Then there are the \textit{complex numbers}, such as:
\[i=\sqrt{-1}, \ \ \ 3+2i, \ \ \ \frac{2}{3}+i\frac{1}{3}.\]
It is important to bear in mind what kind of numbers we are working with. For example,
when we discuss factoring, we often assume that we are working with just the rational numbers.
So we say that \(x^2-4=(x-2)(x+2)\) factors, while \(x^2-3\) and \(x^2+4\) do not factor.
But if we use irrational numbers, then \(x^2-3\) factors just as easily:
\[x^2-3= (x-\sqrt{3})(x+\sqrt{3}),\]
and if we use complex numbers \(x^2+4\) factors as well:
\[x^2+4=(x+2i)(x-2i),\]
as can be checked by expanding the right side.
Remember from the Factor Theorem in Unit 2.6 that if \(x-c\) is a factor of a polynomial
\(f(x)\), then \(f(c)=0\). In other words, \(c\) is a zero of \(f(x)\). So being able to factor
a polynomial is really the same as finding the zeros.
Of course if we had to use irrational or complex numbers to factor the polynomial, then
we can expect the zeros to also be irrational or complex. So the zeros of \(x^2-3\) are
\(\pm\sqrt{3}\) (irrational) and the zeros of \(x^2+4\) are \(\pm 2i\) (complex).
In this course, we will not make use of complex numbers. So we will say that a polynomial
such as \(x^2+4\) cannot be factored. But we will use irrational numbers (as sometimes given
by the quadratic formula) to factor \(x^2-3=(x-\sqrt{3})(x+\sqrt{3})\).
In the case of quadratic functions, if the polynomial has whole numbers as coefficients
and it does not factor using whole numbers, then the quadratic formula will always
give zeros that are irrational or complex numbers.
\subsection{The rational zero theorem}
As we have already mentioned, finding \textbf{all} zeros for a polynomial of degree 3 or higher
can be quite difficult. But in this unit we will focus on how to find the \textbf{rational} zeros.
It turns out that this problem is much easier to solve.
Of course, this does not solve the problem of finding all the zeros, because lots of polynomials
have irrational or complex zeros, but it often helps to find at
least some of the zeros.
We will denote a generic rational number by \(\displaystyle \frac{p}{q}\), where \(p\) and \(q\)
are whole numbers. In case the fraction is negative, we take \(p\) to be negative. For example,
for the fraction \(\displaystyle -\frac{3}{2}\) we use \(p=-3\) and \(q=2\).
So the denominator \(q\) will always be a positive number.
So here is what the Rational Zero Theorem says:
| Rational Zero Theorem |
|---|
| Suppose \(\displaystyle \frac{p}{q} \) is a rational zero of the polynomial
\(f(x)\). |
| Then \(p\) must be a factor of the constant term,
and \(q\) must be a factor of
the leading coefficient.
|
Let's see how this works in an example.
\begin{example}
We are given the polynomial
\[f(x)=2x^3 - 9x^2 + 7x + 6.\]
We would like to find all its rational
zeros (if any). The constant term and leading coefficient are:
\[\mbox{CT}=6, \hspace{2ex} \mbox{LC}=2.\]
The Rational Zero Theorem tells us that if \(p/q\) is a zero, then \(p\) must be a factor
of \(6\) and \(q\) must be a factor of \(2\). The factors of \(6\) are \(1\), \(2\), \(3\) and
\(6\).
But we need to also include the possibility that the fraction \(p/q\) is negative, so \(p\)
may be negative. So the list of possible values for \(p\) is
\[p=\pm 1,\hspace{1ex} \pm 2, \hspace{1ex} \pm 3, \hspace{1ex} \pm 6.\]
The factors of \(2\) are \(1\) and \(2\). So the list of possible values for \(q\)
is:
\[q=1, \hspace{1ex}2\]
Now make the list of all possible fractions with numerator \(p\) and denominator \(q\):
\[\frac{\pm 1}{1}, \hspace{1ex} \frac{\pm 2}{1}, \hspace{1ex} \frac{\pm 3}{1}, \hspace{1ex} \frac{\pm 6}{1} \hspace{1ex}
\frac{\pm 1}{2}, \hspace{1ex} \frac{\pm 2}{2}, \hspace{1ex}
\frac{\pm 3}{2} \hspace{1ex} \frac{\pm 6}{2}.\]
After simplifying and eliminating duplicates, we find the list of all possible rational
zeros of \(f(x)\):
\[\pm 1, \hspace{1ex}\pm 2, \hspace{1ex}\pm 3, \hspace{1ex}\pm 6, \hspace{1ex}\pm \frac{1}{2}, \hspace{1ex}\pm \frac{3}{2}.\]
This is great progress: we have reduced the search for the rational solutions
of \(f(x)=0\) from the infinitely many possibilities of all the rational numbers to
just a few numbers. Note that \(1\) and \(-1\) will always be on the list, because \(1\)
is a factor of any whole number.
Of course not all these eight numbers will be zeros (we know in advance that a polynomial
of degree \(3\) can have at most \(3\) zeros). And in fact it could well be that \textbf{none}
of them is a zero.
To find out, we just need to check. But instead of
substituting the numbers in the formula for \(f(x)\), we will use synthetic division
and the Remainder Theorem. Remember that this is often a more efficient way to do it.
In this case, where we want to try several numbers for the same polynomial, it can be made
even more efficient by using a single table for all the numbers.
We will now begin doing the synthetic division for the polynomial
\(f(x)=2x^3 - 9x^2 + 7x + 6\) and the numbers in the list. Always
start with \(1\), that is guaranteed to be on the list.
We begin by setting up the table as before, but without drawing the horizontal line, and
writing \(1\) on the second line, left of the vertical bar, that we make longer:
| |
\(2\) | \(-9\) | \(7\) |
\(6\) |
| \(1\) | | | | |
| | | | | |
| | | | | |
| | | | | |
We bring the first coefficient down, as before:
| |
\(2\) | \(-9\) | \(7\) |
\(6\) |
| \(1\) |
\(2\) | | |
|
| | |
| | |
| | |
| | |
| | |
| | |
The next step would be to multiply the \(2\) by the \(1\) on the left and write it under the second
coefficient at the top, and the third step would be to add the two numbers. But we combine
these two steps into one, do the operation \(-9+2=-7\) in our head, and write the answer as the second
number on the bottom line. So the \(-7\) in the table below is obtained by multiplying
together the two numbers in bold, and then adding the \(-9\) at the top:
| |
\(2\) | \(-9\) | \(7\) |
\(6\) |
| \(\mathbf{1}\) |
\(\mathbf{2}\) | \(-7\) | |
|
| | |
| | |
| | |
| | |
| | |
| | |
Now we repeat: we multiply the \(-7\) by \(1\) on the left, and add \(7\), and the result
\((-7\times 1)+7=0\) is written at the bottom:
| |
\(2\) | \(-9\) | \(7\) |
\(6\) |
| \(1\) |
\(2\) | \(-7\) | \(0\) |
|
| | |
| | |
| | |
| | |
| | |
| | |
We do one more step, multiplying the \(0\) by \(1\) and adding \(6\): \(0\times 1 +6=6\)
| |
\(2\) | \(-9\) | \(7\) |
\(6\) |
| \(1\) |
\(2\) | \(-7 \) | \(0 \) |
\(6 \) |
| | |
| | |
| | |
| | |
| | |
| | |
The synthetic division of \(f(x)=2x^3 - 9x^2 + 7x + 6\) by \(x-1\) is finished, and the remainder
is not zero (it's \(6\)). This means that \(1\) is not a zero of \(f(x)\), and our first
attempt at finding a rational zero failed.
We write NO to the right, and cross out the entire line.
_______________________
| |
\(2\) | \(-9\) | \(7\) |
\(6\) |
| \(1\) |
\(2\) | \(-7\) | \(0\) |
\(6\) | NO |
| | |
| | |
| | |
| | |
| | |
| | |
We then move to the second possibility: \(x=-1\) (always try \(1\) first and \(-1\) second).
We repeat the synthetic division by writing \(-1\) on the left in the next line, and bring
down the first number at the top, and continue as before:
______________________
| |
\(2\) | \(-9\) | \(7\) |
\(6\) |
| \(1\) |
\(2\) | \(-7\) | \(0\) |
\(6\) | NO |
| \(-1\) | \(2\) |
\( -11\) | \(18 \) | \(-12 \) |
| | |
| | |
| | |
| | |
Once again, we find that the remainder is not zero (it's \(-12\)). So we write another NO on the
right and cross out the entire line:
______________________
______________________
| |
\(2\) | \(-9\) | \(7\) |
\(6\) |
| \(1\) |
\(2\) | \(-7\) | \(0\) |
\(6\) | NO |
| \(-1\) | \(2\) |
\(-11\) | \(18\) | \(-12\) |
NO |
| | |
| | |
| | |
| | |
The next number to try is \(x=2\):
______________________
______________________
| |
\(2\) | \(-9\) | \(7\) |
\(6\) |
| \(1\) |
\(2\) | \(-7\) | \(0\) |
\(6\) | NO |
| \(-1\) | \(2\) |
\(-11\) | \(18\) | \(-12\) |
NO |
| \(2\) | \(2\) |
\(-5 \) | \( -3\) | \( 0\) |
|
| | |
| | |
This time we find that the remainder is \(0\). This is success: we found a zero of \(f(x)\), and
it's \(x=2\). This also means that \(x-2\) is a factor. We write YES together with the factor \((x-2)\)
on the right.
______________________
______________________
| |
\(2\) | \(-9\) | \(7\) |
\(6\) |
| \(1\) |
\(2\) | \(-7\) | \(0\) |
\(6\) | NO |
| \(-1\) | \(2\) |
\(-11\) | \(18\) | \(-12\) |
NO |
| \(2\) | \(2\) |
\(-5\) | \(-3\) | \(0\) |
YES \(x-2\) is a factor |
| | |
| | |
The quotient polynomial is
given by the numbers on the last line:
\[q(x)=2x^2-5x-3.\]
So we can write:
\[2x^3 -9x^2 +7x +6=(x-2)(2x^2-5x-3).\]
After finding the first zero, the quotient polynomial will always be of degree one less than
the original polynomial. Since we started with a degree \(3\) polynomial, the quotient
has degree \(2\).
To find the other zeros, we need to solve \(2x^2-5x-3=0\).
This quadratic polynomial factors as
\[2x^2-5x-3=(2x+1)(x-2)\]
(check!). So we have now completely factored the original polynomial:
\[2x^3 -9x^2 +7x +6=(x-3)(2x+1)(x-2),\]
and we have found all its zeros: \[x=3, x=-\frac{1}{2}, x=2.\]
\end{example}
Our next example will be a polynomial of degree \(4\).
\begin{example}
Let \[f(x)=3x^4 - 5x^3 - 13x^2 + x + 6.\] So
\[\mbox{CT}=6, \hspace{2ex} \mbox{LC}=3,\]
and if \(p/q\) is a rational zeros, the possible values for \(p\) are
\[p=\pm 1,\hspace{1ex} \pm 2,\hspace{1ex} \pm 3,\hspace{1ex} \pm 6.\]
The possible values for \(q\) are:
\[q=\pm 1,\hspace{1ex} \pm 3.\]
So the list of possible values for \(p/q\) is
\begin{eqnarray*}
\frac{p}{q}&=& \frac{\pm 1}{1},\hspace{1ex} \frac{\pm 2}{1},\hspace{1ex} \frac{\pm 3}{1},\hspace{1ex} \frac{\pm 6}{1},\hspace{1ex}
\frac{\pm 1}{3},\hspace{1ex} \frac{\pm 2}{3},\hspace{1ex} \frac{\pm 3}{3},\hspace{1ex} \frac{\pm 6}{3}\\
&=& \pm 1,\hspace{1ex} \pm 2,\hspace{1ex} \pm 3,\hspace{1ex} \pm 6,\hspace{1ex} \pm\frac{1}{3},\hspace{1ex}\pm \frac{2}{3}.
\end{eqnarray*}
We always begin by trying \(x=1\):
| |
\(3\) | \(-5\) | \(-13\) |
\(1\) | \(6\) |
| \(1\) |
\(3\) | \(-2\) | \(-15\) |
\(-14\) | \(-8\) |
| | |
| | |
|
| | |
| | |
|
| | |
| | |
The remainder is not zero. We write NO and cross out the line.
___________________________
| |
\(3\) | \(-5\) | \(-13\) |
\(1\) | \(6\) |
| \(1\) |
\(3\) | \(-2\) | \(-15\) |
\(-14\) | \(-8\) | NO |
| | |
| | |
|
| | |
| | |
|
| | |
| | |
Next we try \(x=-1\):
___________________________
| |
\(3\) | \(-5\) | \(-13\) |
\(1\) | \(6\) |
| \(1\) |
\(3\) | \(-2\) | \(-15\) |
\(-14\) | \(-8\) | NO |
| \(-1\) | \(3\) |
\(-8\) | \(-5\) | \(6\) |
\(0\) |
| | |
| | |
|
| | |
| | |
We found remainder zero. So \(x=-1\) is a zero, and \((x+1)\) is a factor. We write YES
\((x+1)\) on the right.
_____________________________
| |
\(3\) | \(-5\) | \(-13\) |
\(1\) | \(6\) |
| \(1\) |
\(3\) | \(-2\) | \(-15\) |
\(-14\) | \(-8\) | NO |
| \(-1\) | \(3\) |
\(-8\) | \(-5\) | \(6\) |
\(0\) | YES | \((x+1)\) |
| | |
| | |
|
| | |
| | |
Let's stop and think about what we have done so far. We have found a zero \(x=-1\) and a factor
\((x+1)\) for the \(4\)-th degree polynomial \(3x^4 - 5x^3 - 13x^2 + x + 6\). The quotient is
of degree \(3\), and given by the last row in the table:
\[q(x)=3x^3-8x^2-5x+6.\]
So we can write
\[3x^4 - 5x^3 - 13x^2 + x + 6=(x+1)(3x^3-8x^2-5x+6).\]
To continue in our search for the remaining zeros, we now need to focus on the
quotient \(3x^3-8x^2-5x+6\). This means that we no longer need the top line of the table.
So we cross it out, and continue working using the bottom line:
___________________________
___________________________
| |
\(3\) | \(-5\) | \(-13\) |
\(1\) | \(6\) |
| \(1\) |
\(3\) | \(-2\) | \(-15\) |
\(-14\) | \(-8\) | NO |
| \(-1\) | \(3\) |
\(-8\) | \(-5\) | \(6\) |
\(0\) | YES | \((x+1)\) |
| | |
| | |
|
| | |
| | |
It is possible that \((x+1)\) will also be a factor of the quotient
\(3x^3-8x^2-5x+6\). This would mean that \((x+1)\) is a factor of the original
polynomial twice, or in other words that \((x+1)^2\) is a factor. To check this possibility
we need to try the possible zero \(x=-1\) again using the quotient polynomial:
___________________________
___________________________
| |
\(3\) | \(-5\) | \(-13\) |
\(1\) | \(6\) |
| \(1\) |
\(3\) | \(-2\) | \(-15\) |
\(-14\) | \(-8\) | NO |
| \(-1\) | \(3\) |
\(-8\) | \(-5\) | \(6\) |
\(0\) | YES | \((x+1)\) |
| \(-1\) | \(3\) |
\(-11\) | \(6\) | \(0\) |
|
| | |
| | |
We found the remainder to be zero. So \((x+1)^2\) will be a factor of \(f(x)\).
We write another YES \((x+1)\) on the right, and the quotient
polynomial \(3x^2-11x+6\) at the bottom is now of degree \(2\),
so we no longer need the synthetic division:
___________________________
___________________________
| |
\(3\) | \(-5\) | \(-13\) |
\(1\) | \(6\) |
| \(1\) |
\(3\) | \(-2\) | \(-15\) |
\(-14\) | \(-8\) | NO |
| \(-1\) | \(3\) |
\(-8\) | \(-5\) | \(6\) |
\(0\) | YES | \((x+1)\) |
| \(-1\) | \(3\) |
\(-11\) | \(6\) | \(0\) |
YES | \((x+1)\) |
| | |
| | |
The quotient polynomial can be factored:
\[3x^2-11x+6=(3x-2)(x-3).\]
So the zeros of \(f(x)\) are
\[-1, \hspace{1ex} \frac{2}{3}, \hspace{1ex} 3,\]
and we have completely factored \(f(x)\) as
\[3x^4 - 5x^3 - 13x^2 + x + 6=(x+1)^2(3x-2)(x-3).\]
\end{example}
We work one more example that has one rational zero, and two irrational zeros.
\begin{example}
We are given the polynomial
\[f(x)=2x^3 + 3x^2 - 12x - 4.\]
The constant term and leading coefficient are:
\[\mbox{CT}=-4, \hspace{2ex} \mbox{LC}=2.\]
The factors of \(2\) are \(1\) and \(2\). We find the list of all possible rational
zeros of \(f(x)\):
\[\pm 1, \hspace{1ex}\pm 2, \hspace{1ex}\pm 4, \hspace{1ex}\pm \frac{1}{2}.\]
We begin the synthetic division with \(x=1\) as usual:
_______________________
| |
\(2\) | \(3\) | \(-12\) |
\(-4\) |
| \(1\) |
\(2\) | \(5\) | \(-7\) |
\(-11\) | NO |
| | |
| | |
| | |
| | |
| | |
| | |
Next we try \(x=-1\):
______________________
______________________
| |
\(2\) | \(3\) | \(-12\) |
\(-4\) |
| \(1\) |
\(2\) | \(5\) | \(-7\) |
\(-11\) | NO |
| \(-1\) | \(2\) |
\(1\) | \(-13\) | \(9\) |
NO |
| | |
| | |
| | |
| | |
The next number to try is \(x=2\):
______________________
______________________
| |
\(2\) | \(3\) | \(-12\) |
\(-4\) |
| \(1\) |
\(2\) | \(5\) | \(-7\) |
\(-11\) | NO |
| \(-1\) | \(2\) |
\(1\) | \(-13\) | \(9\) |
NO |
| \(2\) | \(2\) |
\(7\) | \(2\) | \(0\) |
YES \(x-2\) is a factor |
| | |
| | |
The quotient polynomial is
\[q(x)=2x^2+7x+2.\]
So we can write:
\[2x^3 + 3x^2 - 12x - 4=(x-2)(2x^2+7x+2).\]
To find the other zeros, we need to solve \(2x^2+7x+2=0\).
In this example, the quadratic polynomial does not factor. But we can find its zeros with the
quadratic formula:
\[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-7\pm \sqrt{7^2-4(2)(2)}}{2(2)}=
\frac{-7\pm\sqrt{33}}{4}.\]
So we have found all the zeros:
\[x=2, \hspace{2ex} x=\frac{-7+\sqrt{33}}{4}, \hspace{2ex} \frac{-7-\sqrt{33}}{4}.\]
Only one zero is rational (\(x=2\)). The other two are irrational.
This means that we can also completely factor the polynomial, because each zero \(x=c\)
corresponds to the factor \( (x-c)\). But we need to make sure that the leading coefficient will
be correct. Each factor \((x-c)\) has leading coefficient \(1\), so if all the
factors are like this, the leading coefficient of the factored polynomial will also be \(1\).
Since our original polynomial has LC\(=2\), we insert a \(2\) in front of all the factors:
\[2x^3 + 3x^2 - 12x - 4=2(x-2)\left( x-\frac{-7+\sqrt{33}}{4}\right)\left( x-\frac{-7-\sqrt{33}}{4}
\right).\]
\end{example}
Problems
\problem
Use the Rational Zero Theorem to find all the possible rational zeros of the polynomials, then
use synthetic division to find the actual rational zeros, and
completely factor the polynomials.
\begin{enumerate}
\item
\(f(x)=2x^4 + x^3 - 17x^2 - 9x - 9\)
\item
\(g(x)=2x^3 - 11x^2 + 9x - 2\)
\item
\(h(x)=x^3+x+1\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
For \(f(x)=2x^4 + x^3 - 17x^2 - 9x - 9\), we have \(\mbox{CT}=-9\),
\(\mbox{LC}=2\). So if \(p/q\) is a rational
zero, then
\[p=\pm 1, \hspace{1ex} \pm 3, \hspace{1ex} \pm 9, \hspace{2ex}
q=1, \hspace{1ex} 2, \hspace{2ex}
\frac{p}{q}=\pm 1, \hspace{1ex} \pm 3, \hspace{1ex} \pm 9, \hspace{1ex} \pm \frac{1}{2}, \hspace{1ex}
\pm \frac{3}{2}, \hspace{1ex} \pm \frac{9}{2}.\]
We begin trying \(x=1\) as usual:
| |
\(2\) | \(1\) | \(-17\) |
\(-9\) | \(-9\) |
| \(1\) |
\(2\) | \(3\) | \(-14\) |
\(-23\) | \(-32\) | |
| | |
| | |
| | |
| | |
| | |
| |
| | |
| | |
The remainder is not zero, so we cross out the line and try \(x=-1\):
___________________________
| |
\(2\) | \(1\) | \(-17\) |
\(-9\) | \(-9\) |
| \(1\) |
\(2\) | \(3\) | \(-14\) |
\(-23\) | \(-32\) | NO |
| \(-1\) | \(2\) |
\(-1\) | \(-16\) | \(7\) |
\(-16\) | | |
| | |
| | |
| |
| | |
| | |
That did not work either, so we cross out the line and try the next whole number, \(x=3\):
___________________________
___________________________
| |
\(2\) | \(1\) | \(-17\) |
\(-9\) | \(-9\) |
| \(1\) |
\(2\) | \(3\) | \(-14\) |
\(-23\) | \(-32\) | NO |
| \(-1\) | \(2\) |
\(-1\) | \(-16\) | \(7\) |
\(-16\) | NO | |
| \(3\) | \(2\) |
\(7\) | \(4\) | \(3\) |
\(0\) | |
| | |
| | |
This time we are successful: we write YES\((x-3\)) and cross out the top line, that we
no longer need, and focus on the quotient polynomial of degree \(3\) at the bottom,
\(2x^3+7x^2+4x+3\):
___________________________
___________________________
___________________________
| |
\(2\) | \(1\) | \(-17\) |
\(-9\) | \(-9\) |
| \(1\) |
\(2\) | \(3\) | \(-14\) |
\(-23\) | \(-32\) | NO |
| \(-1\) | \(2\) |
\(-1\) | \(-16\) | \(7\) |
\(-16\) | NO | |
| \(3\) | \(2\) |
\(7\) | \(4\) | \(3\) |
\(0\) | YES | \((x-3)\) |
| | |
| | |
Because \((x-3)\) is a factor, we try \(x=3\) again, to see if \((x-3)^2\) is a factor:
___________________________
___________________________
___________________________
| |
\(2\) | \(1\) | \(-17\) |
\(-9\) | \(-9\) |
| \(1\) |
\(2\) | \(3\) | \(-14\) |
\(-23\) | \(-32\) | NO |
| \(-1\) | \(2\) |
\(-1\) | \(-16\) | \(7\) |
\(-16\) | NO | |
| \(3\) | \(2\) |
\(7\) | \(4\) | \(3\) |
\(0\) | YES | \((x-3)\) |
| \(3\) | \(2\) |
\(13\) | \(43\) | \(132\) |
NO |
The remainder is not zero, so we cross out the line and try the next number, \(x=-3\):
___________________________
___________________________
___________________________
_____________________
| |
\(2\) | \(1\) | \(-17\) |
\(-9\) | \(-9\) |
| \(1\) |
\(2\) | \(3\) | \(-14\) |
\(-23\) | \(-32\) | NO |
| \(-1\) | \(2\) |
\(-1\) | \(-16\) | \(7\) |
\(-16\) | NO | |
| \(3\) | \(2\) |
\(7\) | \(4\) | \(3\) |
\(0\) | YES | \((x-3)\) |
| \(3\) | \(2\) |
\(13\) | \(43\) | \(132\) |
NO |
| \(-3\) | \(2\) |
\(1\) | \(1\) | \(0\) |
YES | \((x+3)\) |
We found a second factor, \((x+3)\), and now the quotient \(2x^2+x+1\) is of degree
\(2\), so we no longer need synthetic division. The quotient does not factor. The quadratic
formula gives:
\[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-1\pm \sqrt{1^2-4(2)(1)}}{2(2)}=
\frac{-1\pm\sqrt{-7}}{4},\]
so there are no more real zeros, and the quotient does not factor. So the only real zeros of \(f(x)\)
are \(3\) and \(-3\), and the complete factorization is
\[2x^4 + x^3 - 17x^2 - 9x - 9=(x-3)(x+3)(2x^2+x+1).\]
\item For \(g(x)=2x^3 - 11x^2 + 9x - 2\), we have
\(\mbox{CT}=-2\),
\(\mbox{LC}=2\). So if \(p/q\) is a rational
zero, then
\[p=\pm 1, \hspace{1ex} \pm 2, \hspace{2ex}
q=1, \hspace{1ex} 2, \hspace{2ex}
\frac{p}{q}=\pm 1, \hspace{1ex} \pm 2, \hspace{1ex} \pm \frac{1}{2}.\]
We try \(x=1\):
_____________________
| |
\(2\)
| \(-11\) | \(9\) |
\(-2\) | |
| \(1\) |
\(2\) | \(-9\) | \(0\) |
\(-2\) | NO | |
| | |
| | |
| | |
| | |
| | |
| |
| | |
| | |
Next we try \(x=-1\):
_____________________
_____________________
| |
\(2\)
| \(-11\) | \(9\) |
\(-2\) | |
| \(1\) |
\(2\) | \(-9\) | \(0\) |
\(-2\) | NO | |
| \(-1\) | \(2\) |
\(-13\) | \(22\) | \(-24\) |
NO | | |
| | |
| | |
| |
| | |
| | |
We try \(x=2\):
We try \(x=2\):
_____________________
_____________________
_____________________
| |
\(2\)
| \(-11\) | \(9\) |
\(-2\) | |
| \(1\) |
\(2\) | \(-9\) | \(0\) |
\(-2\) | NO | |
| \(-1\) | \(2\) |
\(-13\) | \(22\) | \(-24\) |
NO | | |
| \(2\) | \(2\) |
\(-7\) | \(-5\) | \(-12\) |
NO | |
| | |
| | |
We try \(x=-2\):
_____________________
_____________________
_____________________
_____________________
| |
\(2\)
| \(-11\) | \(9\) |
\(-2\) | |
| \(1\) |
\(2\) | \(-9\) | \(0\) |
\(-2\) | NO | |
| \(-1\) | \(2\) |
\(-13\) | \(22\) | \(-24\) |
NO | | |
| \(2\) | \(2\) |
\(-7\) | \(-5\) | \(-12\) |
NO | |
| \(-2\) | \(2\) |
\(-15\) | \(39\) | \(-80\) |
NO |
We have run out of whole numbers to try, so we try \(x=1/2\):
_____________________
_____________________
_____________________
_____________________
| |
\(2\)
| \(-11\) | \(9\) |
\(-2\) | |
| \(1\) |
\(2\) | \(-9\) | \(0\) |
\(-2\) | NO | |
| \(-1\) | \(2\) |
\(-13\) | \(22\) | \(-24\) |
NO | | |
| \(2\) | \(2\) |
\(-7\) | \(-5\) | \(-12\) |
NO | |
| \(-2\) | \(2\) |
\(-15\) | \(39\) | \(-80\) |
NO |
| \(1/2\) | \(2\) |
\(-10\) | \(4\) | \(0\) |
YES | \( (x-1/2)\)
|
So \(x=1/2\) is a zero. The quotient \(2x^2-10x+4\) of degree \(2\)
cannot be factored other than \(2(x^2-5x+2)\), so we use the quadratic formula to solve
\(x^2-5x+2=0\):
\[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-5)\pm \sqrt{(-5)^2-4(1)(2)}}{2(1)}=
\frac{5\pm\sqrt{17}}{2}.\]
So there are two irrational zeros:
\[\frac{5+\sqrt{17}}{2},\hspace{1ex} \frac{5-\sqrt{17}}{2}.\]
The complete factorization of \(g(x)\) is (remember that we need to make sure the leading coefficient
is correct:
\[2x^3 - 11x^2 + 9x - 2=2\left(x-\frac{1}{2}\right)\left(x-\frac{5+\sqrt{17}}{2}\right)
\left(x-\frac{5-\sqrt{17}}{2}\right).\]
By distributing the coefficient \(2\) inside the first factor, we can also re-write this as:
\[2x^3 - 11x^2 + 9x - 2=(2x-1)\left(x-\frac{5+\sqrt{17}}{2}\right)
\left(x-\frac{5-\sqrt{17}}{2}\right).\]
\item
For \(h(x)=x^3+x+1\), the constant term and leading coefficient are both \(1\). This means that
the only possible rational zeros are \(1\) or \(-1\). In this case, substituting \(1\) or
\(-1\) in the polynomial is the fastest way to check:
\[h(1)=1^3+1+1=3,\\ h(-1)=(-1)^3+(-1)+1=-1-1+1=-1\]
So neither possibility is a zero, and \(h(x)\) has no rational zeros, and it cannot be factored.
But we can say something more about this polynomial. The degree is odd
and the leading coefficient is positive. So we know from Unit 2.4 that the end behavior is of
type Up-Up: \(\img{UU.png}{-1em}{}{2em}\).
This means that there must be at least one
\(x\)-intercept, and so at least one zero. But this zero must be irrational, and it is
difficult to find its value.
\end{enumerate}
\end{sol}
\mproblem
Use the Rational Zero Theorem to find all the possible rational zeros of the polynomials, then
use synthetic division to find the actual rational zeros, and
completely factor the polynomials.
\begin{enumerate}
\item
\(f(x)=3x^3 - x^2 + 6x - 2\)
\item
\(g(x)= x^5 + 4x^4 + 4x^3 + 5x^2 + 7x + 3\)
\item \(h(x)= 6x^4 - 31x^3 + 25x^2 + 39x + 9\)
\end{enumerate}