Transcript x - Barnstable Academy

4.5: More on Zeros of Polynomial Functions
Upper and Lower Bounds for Roots
The Upper and Lower Bound Theorem helps us rule out many of a
polynomial equation's possible rational roots.
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.
4.5: More on Zeros of Polynomial Functions
EXAMPLE: Finding Bounds for the Roots
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.
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4.5: More on Zeros of Polynomial Functions
EXAMPLE: Finding Bounds for the Roots
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 -14
3 0
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.)
4.5: More on Zeros of Polynomial Functions
EXAMPLE: Finding Bounds for the Roots
Show that all the real roots of the equation 8x3 + 10x2 - 39x + 9 = 0 lie
between –3 and 2.
2nd Method For Finding Lower Bound: Similar to Des Carte’s Rule of
signs, we can find f(-x) . Divide the polynomial by x – 3 using synthetic
division (note 3 is now positive). If all of the sings in the bottom row are
negative in value, then -3 is a lower bound. Remember that the number zero
can be considered positive or negative.
3
-8
10
-24
39
9
-42 -9
Counting zero as
negative, the signs
alternate: +, -, +, -.
-8 -14
-3 0
By the Upper and Lower Bound Theorem, the negative 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.) A lower/upper bound may not be a root/zero.
4.5: More on Zeros of Polynomial Functions
EXAMPLE 2: Finding Bounds for the Roots
Use the Upper Bound Theorem to find an integral upper bound and the Lower
Bound to find an integral lower bound of the zeros of
f(x) =x3 + 5x2 - 3x - 20
1.
Determine the number of complex roots
1.
List the possible rational zeros.
1.
Find Possible positive and negative real zeros
1.
Determine interval where the zeros are located
1.
Determine Upper and Lower Bounds
1.
Determine zeros to the nearest tenth.
3.5: More on Zeros of Polynomial Functions
The Intermediate Value Theorem
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.
3.5: More on Zeros of Polynomial Functions
EXAMPLE: Approximating a Real Zero
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 - 5 = 8 - 4 - 5 = -1
and
f (2) is negative.
f (3) is positive.
f(3) = 33 - 2
3 - 5 = 27 - 6 - 5 = 16.
This sign change shows that the polynomial function has a real zero
between 2 and 3.
3.5: More on Zeros of Polynomial Functions
EXAMPLE: Approximating a Real Zero
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
f(x) = x3 - 2x - 5
2
f(2) = 23 - 2(2) - 5
2.1
= -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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3.5: More on Zeros of Polynomial Functions
EXAMPLE: Approximating a Real Zero
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.
3.5: More on Zeros of Polynomial Functions
The Fundamental Theorem of Algebra
We have seen that if a polynomial equation is of degree n, then counting
multiple roots separately, the equation has n roots. This result is called the
Fundamental Theorem of Algebra.
The Fundamental Theorem of Algebra
If f(x) is a polynomial of degree n, where n  1, then the equation f(x) = 0 has
at least one complex root.
3.5: More on Zeros of Polynomial Functions
The Linear Factorization Theorem
Just as an nth-degree polynomial equation has n roots, an nth-degree
polynomial has n linear factors. This is formally stated as the Linear
Factorization Theorem.
The Linear Factorization Theorem
If f(x) = anxn + an-1xn-1 + … + a1x + a0 b, where n  1 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.
3.5: More on Zeros of Polynomial Functions
EXAMPLE:
Finding a Polynomial
Function with Given Zeros
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)
This is the linear factorization for a fourthdegree polynomial.
= an(x + 2)(x -2)(x - i)(x + i)
Use the given zeros: c1 = -2, c2 = 2, c3 = i,
and, from above, c4 = -i.
= an(x2 - 4)(x2 + i)
Multiply
f(x) = an(x4 - 3x2 - 4)
Complete the multiplication
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3.5: More on Zeros of Polynomial Functions
EXAMPLE:
Finding a Polynomial
Function with Given Zeros
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.