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3.2
LOCATING ZEROS
The man who breaks out into a new era of thought is usually himself still a
prisoner of the old. Even Isaac Newton, who invented the calculus as a
mathematical vehicle for his epoch-making discoveries in physics and
astronomy, preferred to express himself in archaic geometrical terms.
Freeman Dyson
I had a good teacher for
freshman algebra. I think
he was simultaneously the
football coach. Then I took
sophomore geometry. It
was apparently thought
that students couldn’t learn
geometry in one year so
they had a second course
in the junior year. The
teacher in this second
course didn’t understand
the subject and I did. I
made a lot of trouble for
her.
Saunders MacLane
In Section 3.1 we indicated that two of the important concerns for polynomials are
locating zeros and local extrema. With calculator graphs we can make excellent
approximations for both. On a graph, there is no obvious relation between zeros
and turning points, but in calculus we learn that every turning point of a polynomial
function f occurs at a zero of another polynomial function called the derivative of
f. Thus the location of turning points also depends on locating zeros. We leave the
study of derivatives to calculus, but we devote this section to understanding more
about zeros of polynomial functions.
It would be nice to have something analogous to the quadratic formula for
higher degree polynomials. For some polynomial functions, zeros can be expressed
in exact form using radicals and the ordinary operations of algebra, but in general,
we must rely on approximations. See the two Historical Notes in this section. Some
of the theorems included in this section help us determine whether or not exact form
solutions are available.
Locator Theorem
Graphs of all polynomial functions share some common properties; they are
continuous and smooth, with no corners, breaks, or jumps. The idea of continuity
(no breaks or jumps) is another topic for calculus and subsequent courses. We need
some such theorem because calculator graphs are neither smooth nor continuous.
Every calculator graph is the result of computing lots of discrete function values
(one for each column of pixels). Dot mode shows only the isolated points. How are
we to be confident that there isn’t some break or jump in the graph between two
adjacent pixel columns? In connected mode, a graphing calculator connects different points of the graph by vertical strips, which we know cannot be part of the
graph of any function.
Nevertheless, our eye smooths out calculator graphs. We have come to expect
graphs to be smooth and continuous, and the following theorem supports our
intuition. It says, in effect, that a polynomial function cannot change from positive
to negative without going through 0. The locator theorem is a special case of a
theorem from analysis called the Intermediate Value Theorem.
Locator (sign-change) theorem
Suppose p is a polynomial function and a and b are numbers such that p~a!
and p~b! have opposite signs. The function p has at least one zero between a
and b, or equivalently, the graph of y 5 p~x! crosses the x-axis between
~a, 0! and ~b, 0!.
cEXAMPLE 1 Using the locator theorem
Let p~x! 5 2x 3 2 2x 2 2 3x 1 1.
(a) Use the locator theorem to verify that p has a zero between 0 and 1.
(b) Using a decimal window, graph y 5 p~x! and trace to verify that the zero is
located between 0.2 and 0.3.
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Solution
(.2, .336)
(a) p~0! 5 1 and p~1! 5 22, so there is a sign change and hence a zero between
0 and 1.
(b) Tracing along the graph of p~x! in a decimal window, we can read the coordinates shown in Figure 11. Thus p~0.2! 5 0.336 and p~0.3! 5 20.026. There
is a sign change between 0.2 and 0.3, so there is a zero in the interval. b
(.3, .026)
[– 5, 5] by [– 3.5, 3.5]
p(x) = 2x3 – 2x2 – 3x + 1
FIGURE 11
We could, of course, use the calculator to locate the zero in Example 1 more
precisely. If, for example, we simply zoom in on the point (0.2, 0) and then trace,
we can locate the zero between 0.275 and 0.30. With time and patience we can
locate zeros with as much accuracy as a calculator will display. A closer approximation is 0.2929.
To get zeros in exact form, to move beyond approximations, we need other
tools. The most powerful technique available to us involves factoring and the zeroproduct principle. Unfortunately, in most cases where we need the zeros of a given
polynomial function, there is no dependable procedure for finding even one zero
in exact form, and with polynomial functions of higher degree, even knowing
several zeros may not be enough.
The Division Algorithm
Just as we can divide one integer by another to get an integer part q and a remainder
r, where the remainder must be smaller than the divisor, so we can divide one
polynomial by another. The result of polynomial division is a polynomial part q~x!,
and a remainder r~x! whose degree must be smaller than the degree of the divisor.
In particular, when the divisor is a linear polynomial (of the form x 2 c), then
the remainder is some number r. This result is stated as a theorem known as the
Division Algorithm. Properly, in the statement of the theorem, the degree of the
divisor, d~x!, must be no greater than the degree of the polynomial we are dividing,
p~x!. In our work, we assume a divisor that is either a linear or a quadratic
polynomial.
Division algorithm
If p~x! is a polynomial of degree greater than zero, and d~x! is a polynomial,
then dividing p~x! by d~x! yields unique polynomials q~x! and r~x!, called the
polynomial part and remainder, respectively, such that
p~x! 5 d~x! · q~x! 1 r~x!,
where the degree of r~x! is smaller than the degree of d~x!.
If d~x! 5 x 2 c, then the remainder is a unique number r, such that
p~x! 5 ~x 2 c! · q~x! 1 r.
(1)
To find the polynomial part and remainder for any given pair of polynomials
we use the familiar process of long division. For the special case of a linear divisor,
there is a shortcut called synthetic division. Synthetic division is stressed in traditional courses because it is also used for several different evaluation purposes. With
graphing calculators, however, the convenience of synthetic division does not justify the time required to learn the process. We outline synthetic division in the
following (optional) discussion. You may divide by any method you wish, but we
will do all of our polynomial division by long division.
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HISTORICAL NOTE
IS THERE A CUBIC FORMULA?
The Babylonians could solve some
quadratics nearly four thousand
years ago, as could the ancient
Greeks and Egyptians although
they thought only positive roots
had meaning. In essence, the
quadratic formula has been around
for at least a thousand years.
From at least 1200 A.D. people
have searched for a comparable
formula for cubics. The story of
Although Cardan is known for
who first succeeded, and when, gets
developing the first cubic
formula, credit actually
muddled by conflicting claims. At
belongs to Tartaglia, pictured
least part of the credit belongs to
here.
Scipione del Ferro (ca. 1510). By
about 1540, Tartaglia had learned enough to win a
public contest, solving 30 cubics in 30 days.
x5
Somehow, Cardan got the method from Tartaglia
and published it in 1545, much to Tartaglia’s
dismay. The solution is often called
Cardan’s even though he did credit
Tartaglia.
The methods of this chapter
are much easier to apply, but the
formula from Cardan’s book still
works. Given a cubic of the form
x 3 1 ax 1 b 5 0,
first calculate
A5
SD SD
a
3
3
1
b 2
.
2
Cardan’s solution is given by
Î
3
ÏA 2
b
2
2
Î
3
ÏA 1
b
.
2
Synthetic Division Algorithm (Optional)
Synthetic division is sometimes a convenient method for factoring a polynomial
function. We do not justify the steps, but the procedure is really nothing but a
short-cut method of dividing, using only the coefficients. The steps are outlined in
the following box.
Synthetic division algorithm (for divisors of the form x 2 c)
To divide a polynomial p~x! of degree n by x 2 c:
1. On the top line write c (change sign from x 2 c), followed by all the
coefficients of p~x! in order of decreasing powers of x, including any zero
coefficients.
2. Bring down the leading coefficient, multiply by c, and add the product to
the next coefficient to get the next entry on the bottom line. Repeat,
multiplying the sum by c, and adding the product to the next coefficient,
and continue for all coefficients of p.
3. The first n numbers on the bottom line are the coefficients of q~x!, of
degree n 2 1, and the last number is the remainder r.
We illustrate the synthetic division algorithm with the problem from Example 2: Divide p~x! 5 2x 3 2 2x 2 2 3x 1 1 by x 1 1. You may want to compare
the coefficients in the long division of Example 2 with the numbers in the synthetic
division. Since x 1 1 5 x 2 ~21!, We write 21 on the top line at the left,
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followed by the coefficients of p~x! in order of decreasing powers of x:
c from x 2 c A
21
22 23
1
22
4 21
2 24
1
0
2
q~x! 5 2x 2 2 4x 1 1
r50
B Coefficients of p~x!
B For each entry on
middle line, multiply
bottom entry by 21,
and add.
From the last line we read the coefficients of the polynomial part q~x!, when p~x!
is divided by x 1 1, and the remainder r. Writing p~x! in the form of the division
algorithm,
2x 3 2 2x 2 2 3x 1 1 5 ~x 1 1!~2x 2 2 4x 1 1! 1 0.
cEXAMPLE 2 Using the division algorithm Verify that 21 is a zero of the
polynomial function from Example 1, p~x! 5 2x 3 2 2x 2 2 3x 1 1, and find the
other two zeros in exact form.
Solution
Substituting 21 for x, p~21! 5 0.
Dividing p~x! by x 1 1 yields the following:
2x 2 2 4x 1 1
x 1 1 ) 2x 3 2 2x 2 2 3x 1 1
2x 3 1 2x 2
24x 2 2 3x
24x 2 2 4x
x11
x11
0
Since the remainder is 0, p~x! can be written in factored form as
2x 3 2 2x 2 2 3x 1 1 5 ~x 1 1!~2x 2 2 4x 1 1!.
By the zero-product principle, either x 1 1 5 0 or 2x 2 2 4x 1 1 5 0. Using the
quadratic formula for the second equation, we find that the zeros of p are 21,
2 2 Ï2 2 1 Ï2
, 2 . The second zero is about 0.29289, clearly the number we were
2
“zeroing in on” in Example 1. b
cEXAMPLE 3
The division algorithm again
(a) Show that x 2 1 4 is a factor of the polynomial
(b) Find all zeros of p~x!.
p~x! 5 x 4 2 x 3 1 2x 2 2 4x 2 8.
Solution
(a) We use long division.
x2 2 x 2 2
x 1 4 ) x 4 2 x 3 1 2x 2 2 4x 2 8
1 4x 2
x4
3
2 x 2 2x 2 2 4x
2 4x
2 x3
28
2 2x 2
28
2 2x 2
0
2
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By the division algorithm, since the remainder is 0,
p~x! 5 ~x 2 1 4!~x 2 2 x 2 2!,
and we have a factorization of p~x!, thus reducing the problem of finding the
zeros of a fourth polynomial function p to solving two quadratic equations,
x 2 1 4 5 0, and x 2 2 x 2 2 5 0.
(b) The four zeros of p~x! are 6 2i, 21, and 2. With two real zeros, the graph of
y 5 p~x! should cross the x-axis just twice. See Figure 12. b
Factor and Remainder Theorems
[– 2, 3] by [– 10, 2]
p(x) = x 4 – x 3 + 2x 2 – 4x – 8
FIGURE 12
Examples 2 and 3 illustrate one of the key concepts in finding exact form zeros of
polynomial functions. Finding a zero is equivalent to finding a factor, and once we
have factors, by the zero-product principle, the zeros of p are the zeros of the
factors. Further, since the degree of q is smaller than the degree of p, q~x! is called
the reduced polynomial.
In the case where the divisor is linear, the division algorithm provides two
powerful theorems. Consider again Equation (1),
p~x! 5 ~x 2 c!q~x! 1 r.
First, suppose r 5 0. Then ~x 2 c! is a factor, p~x! 5 ~x 2 c!q~x!. Conversely, if
~x 2 c! is a factor of p~x!, then p~x! 5 ~x 2 c!q~x!, so r must be 0.
Equation (1) is an identity, so we can replace x by any number and obtain a
true statement. In particular, if we replace x by c, we obtain
p~c! 5 ~c 2 c!q~c! 1 r 5 0 · q~c! 1 r 5 0 1 r 5 r.
Thus the remainder always equals the value of the function p at the number c.
Putting these observations together, we have the following.
Remainder and factor theorems
When p~x! is divided by x 2 c, the remainder is p~c!.
When p~x! is divided by x 2 c, then x 2 c is a factor of p~x! if and only if
p~c! 5 0.
The factor and remainder theorems give us ways to find a remainder without
performing a lengthy division and can simplify many calculations.
cEXAMPLE 4
Strategy: By the remainder
theorem, for (a) p~21! 5 r,
and for (b) P~22! 5 0,
from which we can solve
for k.
The remainder theorem
(a) Find the remainder when the polynomial p~x! 5 4x 15 1 5x 7 1 2x 4 1 3 is
divided by x 1 1.
(b) Find the value of k such that if the polynomial P~x! 5 x 3 1 x 2 1 kx 2 4 is
divided by x 1 2, then the remainder is 0.
Solution
(a) Following the strategy, p~21! 5 4~21! 1 5~21! 1 2~1! 1 3 5 24. Hence,
p~21! 5 24, so r 5 24.
(b) Evaluating P at 22, we have P~22! 5 28 1 4 2 2k 2 4 5 2822k. Since
the remainder when P~x! is divided by x 1 2 is 0, then P~22! 5 0. Thus,
28 2 2k 5 0,
or
22k 5 8,
or k 5 24.
The desired function is
P~x! 5 x 3 1 x 2 2 4x 2 4.
b
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THERE IS NO “QUINTIC FORMULA”
Very shortly after discovery of
the general solution of the cubic
equation (see “Is There a Cubic
Formula?”), Ferrari (Italy, ca.
1545) derived a method for quartics
(polynomials of degree 4). For
n 5 2, 3, or 4, solutions for equations of degree n involve nth roots.
Why not for degree 5? Nearly three
hundred years passed before much
more was done. Then, within a few
years, two brilliant young men completely
resolved the question.
In 1820 Niels Henrik Abel of Norway was 18
when he thought he had the desired formula.
Before it could be checked by others, however, he
found his error and proved that there could be no
general solution for quintic equations.
In Paris in 1829 another
18-year-old, Evariste Galois, took
the final step. In papers written
during 1829 and 1830, Galois found
the conditions that determine just
which polynomial equations of
degree 5 or higher can be solved in
terms of their coefficients.
Abel died of tuberculosis in
1829 at age 26. In 1831, at the age
of 20, Galois was killed in a duel he
himself recognized as stupid. During their brief
careers, they laid the foundations for modern
group theory, which has applications as diverse as
solutions for Rubik’s cube and the standard model
of elementary particle physics at the beginning of
the universe.
Clearing Fractions and Rational Zeros
Multiplying an equation by a nonzero constant to clear fractions does not change
the roots of the equation. For example, multiplying both sides of the equation
2
7
7
x 3 1 x 2 1 x 2 5 0,
3
2
3
by 6, we get an equation with integer coefficients having the same roots,
6x 3 1 21x 2 1 14x 2 4 5 0.
If all coefficients are integers, then the following theorem provides a complete
list of all the rational numbers that can possibly be zeros.
Rational zeros theorem
Let p be any polynomial function with integer coefficients. The only rational
numbers that can possibly be zeros of p are the numbers of the form rs ,
where r is a divisor of the constant term, and s is a divisor of the leading
coefficient.
If none of these numbers is a zero, then p has no rational zeros.
The rational zeros theorem is useful and important because it lists all the
possibilities for rational zeros. The theorem does not tell us whether a given
polynomial has any rational zeros at all; many do not. Without graphing technology, the theorem is a great help in guiding the search for zeros. Used with a grapher,
the theorem can tell us things about graphs that the calculator cannot. No calculator can distinguish between rational and irrational numbers; every decimal is
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truncated (“cut off ”) to fit the display capacity. Knowing what the rational possibilities are, we can use a calculator to verify that a particular zero is or is not a rational
number.
cEXAMPLE 5 Rational possibilities Use the rational zeros theorem to list
all possible zeros of the polynomial function.
(a) P~x! 5 x 3 2 4x 2 1 x 2 6
Strategy: (b) To get integer
coefficients, multiply
through by 9 and then apply
the rational zeros theorem.
Zeros of 9R are are the
same as the zeros of R.
(b) R~x! 5 x 4 2 4x 3 1
14
9
x2 1
44
9
x2
5
3
Solution
(a) For P, begin by listing all possible numerators (factors of 26) and denominators (factors of 1):
Possible numerators:
Possible denominators:
6
6
1, 2, 3, 6
1
The only possibilities for rational zeros of P are the integers 26, 23, 22, 21,
1, 2, 3, and 6.
(b) Follow the strategy and find all possible rational zeros of S~x! 5 9R~x!:
S~x! 5 9x 4 2 36x 3 1 14x 2 1 44x 2 15.
Possible numerators are factors of 15; denominators are factors of 9.
Possible numerators:
Possible denominators:
6 1, 3, 5, 15
6 1, 3, 9
The rational zeros theorem tells us that S, and hence R, has only sixteen possible rational zeros (in reduced form):
F
6 1, 3, 5, 15, 13 , 53 , 19 ,
5
9
G
. b
Having listed lots of possibilities for rational zeros of the polynomials P and R
in Example 5, what do we know of the actual zeros? At this stage, we have nothing
but possibilities. When we add graphing technology, we can say a great deal more,
as in the next example.
cEXAMPLE 6 Finding rational zeros Find all rational zeros, if there are
any, of the polynomial functions P and R in Example 5. Approximate any irrational
zeros to two decimal place accuracy.
(a) P~x! 5 x 3 2 4x 2 1 x 2 6
(b) R~x! 5 x 4 2 4x 3 1
14
9
x2 1
44
9
x2
5
3
Solution
(a) We don’t know anything about P except that as a cubic it must have at least one
real zero. We begin with a graph in the @210, 10# 3 @210, 10# window. See
Figure 13a. It is clear that there is exactly one real zero, very near 4. Looking
at the list of rational zeros for P from Example 5, we see that the only positive
rational possibilities are 1, 3, and 6. Therefore, the real zero of the polynomial
P cannot be a rational number. If we zoom into a very small box around the
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8
4
–8
–4
4
8
4 4.05 4.10 4.15
–4
–8
[3.8, 4.2] by [– 0.1, 0.1]
(b)
[– 10, 10] by [– 10, 10]
(a)
FIGURE 13
P~x! 5 x 3 2 4x 2 1 x 2 6
x-intercept point of the graph (Figure 13b) and then trace, we find that the zero
is very near 4.11.
(b) For graphing purposes, we can choose either the polynomial R~x!, or the polynomial S~x! 5 9R~x! with integer coefficients, because R and S have the same
zeros. If we were working by hand, most of us would choose S to avoid
fractions; for the calculator there is no difference, except that R requires a
smaller window. The graph of R is shown in the @210, 10# 3 @210, 10#
window we used for part (a) in Figure 14a. When we zoom into a box just large
enough to include the zeros of the graph, as in Figure 14b, we can trace along
the curve to find zeros near 21, 0.3, 1.6, and 3.
–1
[– 10, 10] by [– 10, 10]
(a)
1
2
3
[– 1.5, 4] by [– 3.2, 3]
(b)
FIGURE 14
Which, if any, of these are rational zeros? The list of rational possibilities (from
Example 5) includes 21, 3, 13 , and 53 , all of which are reasonable candidates, but
are these actually zeros of the function? To decide, we need to evaluate R at each
number (see the following Technology Tip). To calculator accuracy we find that
R~21! 5 0,
R~3! 5 0,
R~13 ! 5 0,
R~53 ! 5 0.
We conclude that R has four rational zeros, 21, 3, 13 , and 53 . b
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TECHNOLOGY TIP r Function evaluation
Most calculators, when we trace along a curve, display coordinates. The
y-coordinate is the calculated value corresponding to the x-coordinate of the
pixel, but we have no way to specify a particular x-value unless our window
happens to have it as a pixel. If our goal is to evaluate R~ 13 !, we don’t want to
settle for R (.324076113). We give here some suggestions for different
calculators, but there may be a more efficient way for your particular
machine. The displayed value is the calculator’s evaluation of R~ 13 !, which
may involve round-off error. For example, one of our calculators displays
R~ 13 !, 5 3E-13, meaning 0.0000000000003, and which in this context we
interpret as 0.
TI calculators: If you are graphing a function as Y1, return to the home
screen, store the desired value in the x-register, and then enter Y1 . Thus, for
R~21!, 21A X Enter. Then call up Y1 from the Y-vars menu (or on the TI-85,
2nd Alpha y1, and Enter. The TI-82 will evaluate Y1(21) directly.
Casio calculators: The function must be entered on your MEM list, so type
in the function, SHIFT MEM, F1(STO) and the number, say 1 for f1 .To evaluate f1~3!,
EXIT and store 3 in the x-register, 3 A X EXE . Then F2(RCL) 1 EXE.
HP–38: Having entered a function as, say, F1~X!, return to the home
screen, type F1~1y3!, and ENTER.
HP-48: The calculator will evaluate the function at any pixel-address
and store the result on the stack, but direct evaluation is less convenient. One
way is to store the number in, say, register A: 21 ENT ‘A’ STO. Then write the
function as an expression in A: ‘A ` 4 2 4*A ` 3 1 14/9*A ` 2 1 44/9*A 2 5/3’. ENTER twice, so
you have an extra copy on the stack. Then, purple NUM converts to a number.
For another value, store it in A, Enter, and evaluate as a number.
All these calculators except the TI-81 have some sort of SOLVE routine
that will approximate zeros directly, which does not make it less important
for you to understand the ideas we are discussing here.
Applying the Rational Zeros Theorem
The rational zeros theorem may be applied to problems other than looking for the
zeros of a particular polynomial function. For instance, if we know that some
number c is a zero of a polynomial function but c is not among the possibilities for
rational zeros, then we can conclude that c is not a rational number, as in the next
example.
cEXAMPLE 7
Showing a number is not rational
(a) Find a polynomial equation with integer coefficients satisfied by the number
3
c 5 Ï5 2 1.
(b) Use the rational zeros theorem to show that c is not a rational number.
Solution
(a) The simplest polynomial equation satisfied by c is
3
x 5 Ï5 2 1,
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but of course, this does not have integer coefficients. To get integer coefficients,
we add 1, cube both sides, and simplify: @See the inside front cover for the
expansion of ~x 1 1!3.]
3
~x 1 1!3 5 ~Ï5!3
x 3 1 3x 2 1 3x 1 1 5 5
x 3 1 3x 2 1 3x 2 4 5 0.
Now we have a polynomial equation, one of whose solutions is the number c.
Incidentally, a graph makes it clear that p~x! 5 x 3 1 3x 2 1 3x 2 4 has only
3
one real zero, which thus must be Ï5 2 1.
(b) Since the leading coefficient of p~x! is 1, the only possible rational zeros of p
are 6 @1, 2, 4#. The number c is a zero of p but is not one of the possible
3
rational zeros, so Ï5 2 1 must be an irrational number. b
(.5, 0)
(– 1, 0)
(2, 0)
[– 5, 5] by [– 3.5, 3.5]
2x 3 – 3x 2 – 3x + 2 $ 0
on [– 1, 0.5] < [2, `)
cEXAMPLE 8 Solving an inequality
2x 3 2 3x 2 2 3x 1 2 $ 0.
Find the solution set for
Solution
A graph of f ~x! 5 2x 3 2 3x 2 2 3x 1 2 is shown in Figure 15. It appears from
the graph that the zeros are 21, 12 , and 2, as we may verify, either by tracing in a
decimal window, or by evaluating the function. The function f is clearly positive
between 21 and 12 , and whenever x . 2. Therefore, the solution set for the inequality is given by
FIGURE 15
S 5 @21, 0.5# < @2, `!.
b
EXERCISES 3.2
Check Your Understanding
Draw a graph whenever helpful.
Exercises 1–6 True or False. Give reasons.
1. The function p~x! 5 4x 3 2 x has three real zeros.
2. The positive zero of f ~x! 5 x 3 2 3x is less than 1.73.
3. For f ~x! 5 x 3 2 1.6x 2 2 8.52x 1 15.84, since f ~2!
and f ~3! are positive, then f contains no zeros between
2 and 3.
4. The equation 2x 3 2 5x 2 1 4x 2 1 5 0 has no rational roots.
5. The function f ~x! 5 ~3x 2 2!~x 2 2 2x 2 4! has exactly one real zero.
6. When x 3 2 2x 2 1 3x 2 16 is divided by x 2 3, then
the remainder is 2.
Exercises 7–10 Fill in the blank so that the resulting
statement is true.
7. If x 3 1 2x 2 1 1 5 ~x 1 1!~x 2 1 x 2 1! 1 r for every value of x, then r 5
.
8. The number of rational zeros of
.
f ~x! 5 ~x 2 2 2!~x 2 2 2x 1 3! is
9. The number of real roots of
.
~x 2 2 2!~x 2 2 2x 1 3! 5 0 is
10. If x 37 2 2x 24 1 3x 2 2 5 is divided by x 1 1, then the
remainder is
.
Develop Mastery
Exercises 1–4 Locator Theorem Use the locator theorem to determine which half of the interval contains a zero
of the function.
1. p~x! 5 x 3 2 3x 1 1; @22, 21#
2. f ~x! 5 2x 3 1 3x 2 2 x 2 2; @0,1#
3. g~x! 5 x 3 2 5x 2 1 5x 1 3; @21, 0#
4. p~x! 5 x 3 2 5x 2 1 7x 2 2; @2.5, 3#
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Exercises 5–8 Division Algorithm Use division to find
the polynomial part q~x! and remainder r when p~x! is divided by the given divisor. Write the result in the form
p~x! 5 ~x 2 c!q~x! 1 r, and find p~c!.
5. p~x! 5 2x 3 1 3x 2 2 x 2 2; x 2 1
6. p~x! 5 2x 3 1 3x 2 2 x 2 2; x 1 2
7. p~x! 5 3x 4 1 x 3 2 2x 2 1 x 2 1; x 1 1
8. p~x! 5 x 3 2 3x 2 2 x 1 2; x 2 4
Exercises 9–10 Remainder Find the remainder when
the polynomial is divided by x 2 c.
9. 4x 12 2 3x 8 1 5x 2 2 2x 1 3; c 5 21
10. x 10 2 64x 4 1 3; c 5 2
Exercises 11–14 Factor Theorem Use the factor theorem to find the value of k so that the given linear expression
is a factor of the polynomial.
11. 2x 3 1 4x 2 1 kx 2 3; x 1 2
12. x 4 1 kx 2 1 kx 1 2; x 2 2
13. kx 3 1 3x 2 2 4kx 2 7; x 2 3
14. x 3 2 k 2x 1 ~k 1 1!; x 2 k
Exercises 15–18 Remainder Theorem Use the remainder theorem to find the value of k so that when p~x! is divided
by the linear expression you get the given remainder.
15. p~x! 5 2x 3 1 4x 2 1 kx 2 3; x 1 2; r 5 0
16. p~x! 5 2x 3 1 4x 2 1 kx 1 3; x 1 2; r 5 3
17. p~x! 5 x 3 1 kx 2 2 kx 1 8; x 2 2; r 5 1
18. p~x! 5 x 3 1 kx 2 2 kx 1 8; x 2 2; r 5 0
Exercises 19–24 Rational Zeros Theorem (a) Apply
the Rational Zeros Theorem to list all of the possible rational zeros of f. If the theorem does not apply, explain why.
(b) Use a calculator graph to help you eliminate some to the
numbers listed in part (a).
19. f ~x! 5 6x 3 1 3x 2 2 2x 2 1
20. f ~x! 5 6x 3 2 2x 2 2 9x 1 3
21. f ~x! 5 2x 4 2 2x 3 2 6x 2 1 x 1 2
22. f ~x! 5 6x 3 2 x 2 2 13x 1 8
23. f ~x! 5 3x 3 2 1.5x 2 1 x 2 0.5
24. f ~x! 5 x 3 2 2x 2 1 Ï2x 2 2
Exercises 25–30 Exact Form Zeros (a) Find all zeros
of f (including any complex numbers) in exact form. First
look for rational zeros and express f ~x! in factored form
(linear or quadratic factors). (b) Find the solution set for
p~x! , 0.
25. f ~x! 5 x 3 2 4x 2 1 2x 2 8
26. f ~x! 5 4x 3 2 4x 2 2 19x 1 10
27. f ~x! 5 x 3 2 2.5x 2 2 7x 2 1.5
28. f ~x! 5 x 3 2 3.5x 2 1 0.5x 1 5
29. f ~x! 5 6x 4 2 13x 3 1 2x 2 2 4x 1 15
30. f ~x! 5 4x 4 2 4x 3 2 7x 2 1 4x 1 3
Exercises 31–32 Solving Polynomial Equations
the solution set.
31. (a) 3x 2 2 12x 5 ~x 2 1!~x 2 2 4x!
3x 2 2 12x
5x21
(b)
x 2 2 4x
32. (a) 3x 3 2 12x 5 ~x 1 2!~x 3 2 4x!
3x 3 2 12x
5 x 3 2 4x
(b)
x12
Find
Exercises 33–34 Exact Form Roots (a) Find the roots
in exact form. (Hint: The equation is quadratic in x 2. Use
the quadratic formula to first find x 2.) (b) Get approximations (2 decimal places) to the answers in part (a). Graph as
a check.
34. x 4 2 2x 2 2 1 5 0
33. x 4 2 4x 2 1 1 5 0
Exercises 35–38 Solution Set Find the solution set for
(a) f ~x! 5 0, (b) f ~x 2 1! 5 0, (c) f ~x! # 0. (Hint: For
part (c), first factor and get cut points.)
35. f ~x! 5 2x 3 2 3x 2 2 3x 1 2
36. f ~x! 5 x 4 2 2x 3 2 3x 2 1 4x 1 4
37. f ~x! 5 x 3 2 3x 1 2
38. f ~x! 5 4x 3 2 4x 2 2 19x 1 10
Exercises 39–42 Nonrational Numbers (a) Find a
polynomial equation with integer coefficients having c as a
root. (b) Explain why c is not a rational number. (Hint: See
Example 7.)
39. c 5 Ï2
40. c 5 2 1 Ï5
3
3
42. c 5 2Ï31 1
41. c 5 Ï2 2 1
Exercises 43–46 Verbal to Formula (a) Find a formula
(in expanded form) for a polynomial function satisfying the
given conditions. (b) How many turning points does the
graph have?
43. Degree 3; zeros are 22, 1, and 3; leading coefficient
is 1.
44. Degree 3; zeros are 21, 2, and 4; leading coefficient is
22.
45. Degree 4; zeros are 21, 2, and a double zero at 1; graph
of f contains the point (0, 22).
46. Degree 4; zeros are 0, 2, and ~x 2 2 2x 2 5! is a factor
of f ~x!; graph contains the point (3, 6).
Exercises 47–48 Zeros and Turning Points (a) How
many real zeros does f have? (b) Find all rational zeros.
(c) How many turning points does the graph of f have, if
any? In what quadrants?
47. f ~x! 5 ~x 2 2 4!~x 2 2 8x 1 15!
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48. f ~x! 5 ~2x 2 2 x 2 3!~x 2 1 2x 1 4!
49. Is there a polynomial function of degree 3 that has zeros
at 22, 21, and 2, whose graph passes through the
points (0, 4), (3,220)? Explain.
70. Repeat Exercise 69 for f ~x! 5 x 3 2 cx 1 16.
71. Cone in a Sphere A cone is inscribed in a sphere of
radius 8 cm. See the diagrams showing cross sections.
Let r denote the radius of the cone and h the height.
Exercises 50–53 End Behavior and Local Minima The
intercept points for a polynomial function f of degree 3 are
given. (a) Draw a rough sketch and determine the end behavior. (b) Determine the coordinates (one decimal place)
of the local minimum point.
50. ~22, 0!, ~1, 0!, ~3, 0!, ~0, 6!
51. ~22, 0!, ~1, 0!, ~3, 0!, ~0, 26!
52. ~24, 0!, ~22, 0!, ~1, 0!, ~0, 8!
53. ~25, 0!, ~21, 0!, ~1, 0!, ~0, 210!
h
Exercises 54–57 Local Maxima For the function in Exercises 50–53, give the coordinates (one decimal place) of
the local maximum point.
(a)
Exercises 58–61 Your Choice Suppose the graph of a
polynomial function f of degree 3 has a local maximum
point at P and a local minimum point at Q. Draw a rough
sketch (or sketches) of the graph of f and use it to describe
any features such as the location of x-intercept points, end
behavior, and any other properties. Write an equation for
your function.
58. Both P and Q are in Quadrant I.
59. P is in Quadrant I and Q is in Quadrant IV.
60. P is in Quadrant II and Q is in Quadrant IV.
61. P is in Quadrant I and Q is in Quadrant III.
Exercises 62–65 Bracketing Roots Find the smallest interval with integer endpoints @b, c# containing all the roots
of the equation; that is, b is the largest integer smaller than
all roots of the equation, and similarly for c.
62. 2x 3 2 3x 2 2 2x 1 3 5 0
63. x 3 2 3x 2 2 2x 1 4 5 0
64. x 4 2 2x 2 2 3 5 0
65. x 3 2 3x 2 2 2x 1 8 5 0
66. (a) For what number c is c a zero of
f ~x! 5 2x 3 2 cx 2 1 ~3 2 c 2!x 2 6?
(b) Using your value for c, draw a graph and verify that
c is a zero of f.
Exercises 67–68 Bracketing Zeros Find the smallest interval with integer endpoints @b, c# containing all the zeros
of the function. See Exercises 62–65.
67. f ~x! 5 x 4 2 9x 2 1 6x 2 4
68. f ~x! 5 x 4 2 6x 2 1 3x 1 4
69. Explore Try integer values of c in f ~x! 5 x 3 2
cx 1 2 until you find one for which f has a repeated
zero. Then determine the other zero. Justify your conclusions algebraically.
r
h
8
r
(b)
h
r
8
(c)
(a) Show that for both diagrams, r 2 5 16h 2 h 2. Express the volume V of the cone as a function of h.
(b) Of all such possible cones, there is one that has the
greatest volume. What are the radius and height
(1 decimal place) giving maximum volume? What is
the maximum volume?
72. Cone in a Sphere (Alternative Approach) Solve Exercise 71 by expressing V as a function of r. Explain why
it is not necessary to consider the second diagram to find
the maximum volume.