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Zeros of Polynomial
Functions
The Rational Zero Theorem
• If f (x) = anxn + an-1xn-1 +…+ a1x + a0 has
integer coefficients and p/q (where p/q is
reduced) is a rational zero, then p is a factor
of the constant term a0 and q is a factor of
the leading coefficient an.
Example
Find all of the possible real, rational roots of
f(x) = 2x3-3x2+5
Solution:
p is a factor of 5 = 1, 5
q is a factor of 2 = 1, 2
p/q = 1, 1/2, 5, 5/2
1
Properties of Polynomial
Equations
• If a polynomial equation is of degree n, then
counting multiple roots separately, the
equation has n roots
• If a+bi is a root of the equation, then a-bi is
also a root.
Example
Find all zeros of f(x) = x3+12x2+21x+10
p/q = 1, 2, 5, 10
f(1) = 44
f(-1) = 0
Divide out -1 to get x2+11x-10
Use the quadratic formula to find the last 2
zeros. x=-11.844 and .844
The solutions are -1, -11.844, and .844
Text Example
Solve:
x − 6x2 − 8x + 24 = 0.
4
Solution The graph of f (x) = x4 − 6x2 − 8x + 24 is shown the figure below.
Because the x-intercept is 2, we will test 2 by synthetic division and show that
it is a root of the given equation.
2
•
1
xintercept:
2
0 −6 −8 24
2
4 −4 −24
2 −2 −12
0
The zero remainder
indicates that 2 is a root
of x4 − 6x2 − 8x + 24 = 0.
2
Text Example cont.
Solve:
Solution
x4 − 6x2 − 8x + 24 = 0.
Now we can rewrite the given equation in factored form.
x4 − 6x2 + 8x + 24 = 0
(x – 2)(x3 + 2x2 − 2x − 12) = 0
x–2=0
or
x + 2x − 2x − 12 = 0
3
2
This is the given equation.
This is the result obtained from the
synthetic division.
Set each factor equal to zero.
Text Example cont.
Solve:
x4 − 6x2 − 8x + 24 = 0.
Solution We can use the same approach to look for rational roots of the
polynomial equation x3 + 2x2 − 2x − 12 = 0, listing all possible rational roots.
However, take a second look at the figure of the graph of x4 − 6x2 − 8x + 24 =
0. Because the graph turns around at 2, this means that 2 is a root of even
multiplicity. Thus, 2 must also be a root of x3 + 2x2 − 2x − 12 = 0, confirmed
by the following synthetic division.
These are the coefficients
of x3 + 2x2 − 2x − 12 = 0.
•
1
1
2 −2 −12
2
8
12
4
6
0
The zero remainder
indicates that 2 is a root
of x3 + 2x2 − 2x − 12 = 0.
xintercept:
2
Text Example cont.
Solve:
Solution
x4 − 6x2 − 8x + 24 = 0.
Now we can solve the original equation as follows.
x4 − 6x2 + 8x + 24 = 0
This was obtained from the first
synthetic division.
(x – 2)(x – 2)(x + 4x + 6) = 0
This was obtained from the second
synthetic division.
2
x – 2 = 0 or x – 2 = 0 or x + 4x + 6 = 0
2
x=2
This is the given equation.
(x – 2)(x3 + 2x2 − 2x − 12) = 0
x=2
x2 + 4x + 6 = 0
Set each factor equal to zero.
Solve.
3
Text Example cont.
Solve:
Solution
x4 − 6x2 − 8x + 24 = 0.
We can use the quadratic formula to solve x2 + 4x + 6 = 0.
The solution set of the original equation is: {2, -2 - i√2, -2+i√2}
Descartes’s Rule of Signs
If f (x) = anxn + an−−1xn−−1 + … + a2x2 + a1x + a0 be a polynomial
with real coefficients.
1. The number of positive real zeros of f is either equal to
the number of sign changes of f (x) or is less than that
number by an even integer. If there is only one variation in
sign, there is exactly one positive real zero.
2. The number of negative real zeros of f is either equal to
the number of sign changes of f (−x) or is less than that
number by an even integer. If f (−x) has only one variation
in sign, then f has exactly one negative real zero.
Text Example
Determine the possible number of positive and negative real zeros of
f (x) = x3 + 2x2 + 5x + 4.
Solution
1. To find possibilities for positive real zeros, count the number of sign
changes in the equation for f (x). Because all the terms are positive, there
are no variations in sign. Thus, there are no positive real zeros.
2. To find possibilities for negative real zeros, count the number of sign
changes in the equation for f (−x). We obtain this equation by replacing x
with −x in the given function.
f (x) = x3 + 2x2 + 5x + 4
This is the given polynomial function.
Replace x with −x.
f (−x) = (−x)3 + 2(−x)2 + 5(−x) + 4
= −x3 + 2x2 − 5x + 4
4
•Text Example cont.
Determine the possible number of positive and negative real zeros of
f (x) = x3 + 2x2 + 5x + 4.
Solution
Now count the sign changes.
f (−x) = −x3 + 2x2 − 5x + 4
1
3
2
There are three variations in sign. The number of negative real zeros of f is
either equal to the number of sign changes, 3, or is less than this number by an
even integer. This means that there are either 3 negative real zeros or 3 − 2 = 1
negative real zero.
Zeros of Polynomial
Functions
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The Upper and Lower Bound
Theorem
• Let f (x) be a polynomial with real coefficients and a
positive leading coefficient, and let a and b be nonzero real
numbers.
1. Divide f (x) by x − b (where b > 0) using synthetic
•
division. If the last row containing the quotient and
remainder has no negative numbers, then b is an upper
bound for the real roots of f (x) = 0.
2. Divide f (x) by x − a (where a < 0) using synthetic
•
division. If the last row containing the quotient and
remainder has numbers that alternate in sign (zero entries
count as positive or negative), then a is a lower bound for
the real roots of f (x) = 0.
Text Example
Show that all the real roots of the equation 8x3 + 10x2 − 39x + 9 = 0 lie
between –3 and 2.
Solution We begin by showing that 2 is an upper bound. Divide the
polynomial by x − 2. If all the numbers in the bottom row of the synthetic
division are nonnegative, then 2 is an upper bound .
2
8 10 −39
9
16
52 26
8 26
13 35
All numbers in this row
are nonnegative.
Text Example cont.
Show that all the real roots of the equation 8x3 + 10x2 − 39x + 9 = 0 lie
between –3 and 2.
Solution The nonnegative entries in the last row verify that 2 is an upper
bound. Next, we show that −3 is a lower bound. Divide the polynomial by x −
(−3), or x + 3. If the numbers in the bottom row of the synthetic division
alternate in sign, then −3 is a lower bound. Remember that the number zero
can be considered positive or negative.
−3 8
10 −39
−24
9
42 −9
Counting zero as
negative, the signs
alternate: +, −, +, −.
8
26 13 35
By the Upper and Lower Bound Theorem, the alternating signs in the last row
indicate that −3 is a lower bound for the roots. (The zero remainder indicates
that −3 is also a root.)
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The Intermediate Value Theorem
for Polynomials
• Let f (x) be a polynomial function with real
coefficients. If f (a) and f (b) have opposite
signs, then there is at least one value of c
between a and b for which f (c) = 0.
Equivalently, the equation f (x) = 0 has at
least one real root between a and b.
Text Example
a. Show that the polynomial function f (x) = x3 − 2x − 5 has a real zero
between 2 and 3.
b. Use the Intermediate Value Theorem to find an approximation for this
real zero to the nearest tenth
Solution
a. Let us evaluate f (x) at 2 and 3. If f (2) and f (3) have opposite signs, then
there is a real zero between 2 and 3. Using f (x) = x3 − 2x − 5, we obtain
f (2) = 23 − 2 2 2 − 5 = 8 − 4 − 5 = −1
and
f (2) is negative.
f (3) is positive.
f (3) = 3 − 2 2 3 − 5 = 27 − 6 − 5 = 16.
3
This sign change shows that the polynomial function has a real zero
between 2 and 3.
Text Example cont.
a. Show that the polynomial function f (x) = x3 − 2x − 5 has a real zero
between 2 and 3.
b. Use the Intermediate Value Theorem to find an approximation for this
real zero to the nearest tenth
Solution
b. A numerical approach is to evaluate f at successive tenths between 2 and
3, looking for a sign change. This sign change will place the real zero
between a pair of successive tenths.
x
2
2.1
f(x) = x3 − 2x − 5
f (2) = 23 − 2(2) − 5
= −1
f (2.1) = (2.1)3 − 2(2.1) − 5 = 0.061
Sign change
Sign change
The sign change indicates that f has a real zero between 2 and 2.1.
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Text Example cont.
a. Show that the polynomial function f (x) = x3 − 2x − 5 has a real zero
between 2 and 3.
b. Use the Intermediate Value Theorem to find an approximation for this
real zero to the nearest tenth
Solution
b. We now follow a similar procedure to locate the real zero between
successive hundredths. We divide the interval [2, 2.1] into ten equal subintervals. Then we evaluate f at each endpoint and look for a sign change.
f (2.00) = −1
f (2.04) = −0.590336
f (2.08) = −0.161088
f (2.01) = −0.899399
f (2.05) = −0.484875
f (2.09) = −0.050671
f (2.02) = −0.797592
f (2.06) = −0.378184
f (2.1) = 0.061
f (2.03) = −0.694573
f (2.07) = −0.270257
Sign change
The sign change indicates that f has a real zero between 2.09 and 2.1.
Correct to the nearest tenth, the zero is 2.1.
The Fundamental Theorem of
Algebra
• If f (x) is a polynomial of degree n, where
n ∈ I, then the equation f (x) = 0 has at least
one complex root.
The Linear Factorization
Theorem
If f (x) = anxn + an−−1xn−−1 + … + a1x + a0 b,
where n ∈ I and an ≠ 0 , then
f (x) = an (x − c1) (x − c2) … (x − cn)
where c1, c2,…, cn are complex numbers
(possibly real and not necessarily distinct).
In words: An nth-degree polynomial can be
expressed as the product of n linear factors.
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Text Example
Find a fourth-degree polynomial function f (x) with real
coefficients that has −2, 2, and i as zeros and such that
f (3) = −150.
Solution Because i is a zero and the polynomial has real coefficients, the
conjugate must also be a zero. We can now use the Linear Factorization
Theorem.
f (x) = an(x − c1)(x − c2)(x − c3)(x − c4)
= an(x + 2)(x −2)(x − i)(x + i)
= an(x − 4)(x + i)
2
2
This is the linear factorization for a fourthdegree polynomial.
Use the given zeros: c1 = −2, c2 = 2, c3 = i,
and, from above, c4 = −i.
Multiply
f (x) = an(x4 − 3x2 − 4)
Complete the multiplication
Text Example cont.
Find a fourth-degree polynomial function f (x) with real
coefficients that has −2, 2, and i as zeros and such that
f (3) = −150.
Solution
f (3) = an(34 − 3*32 − 4) = −150
To find an, use the fact that f (3) = −150.
an(81 − 27 − 4) = −150
Solve for an.
50an = −150
an = −3
Substituting −3 for an in the formula for f (x), we obtain
f (x) = −3(x4 − 3x2 − 4).
Equivalently,
f (x) = −3x4 + 9x2 + 12.
Example
Use the roots to find the linear factorization of the
polynomial equation x3-7x2+16x-10
Solution:
The solutions are 3+i, 3-i, and 1
Therefore, x=3+i, x=3-i, and x=1
Getting zero on one side we have the factors (x(3+i)), (x-(3-i)), and (x-1)
The linear factorization is:
(x-3-i)(x-3+i)(x-1)= 0
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