4.5: more on zeros of polynomial functions the upper and lower bound theorem helps us rule out many...

4.5: More on Zeros of Polynomial Functions

The Upper and Lower Bound Theorem helps us rule out many of a polynomial equation's possible rational roots.

The Upper The Upper and and Lower Lower Bound Bound TheoremTheorem Let f (x) be a polynomial with real coefficients and a positive leading coefficient, and let a and b be nonzero real numbers.

1.1. 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.

2.2. 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.

The Upper The Upper and and Lower Lower Bound Bound TheoremTheorem Let f (x) be a polynomial with real coefficients and a positive leading coefficient, and let a and b be nonzero real numbers.

1.1. 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.

2.2. 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.

Upper and Lower Bounds for Roots

EXAMPLE: Finding Bounds for the RootsShow 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 non negative, 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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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 9

24 42 9

8 -14 3 0

Counting zero as negative, the signs alternate: , , , .

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.)



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 39 9

24 -42 9

-8 -14 -3 0

Counting zero as negative, the signs alternate: , , , .

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.


EXAMPLE 2: Finding Bounds for the RootsUse 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.


The Intermediate Value Theorem

The Intermediate Value Theorem for The Intermediate Value Theorem for PolynomialsPolynomialsLet 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.

The Intermediate Value Theorem for The Intermediate Value Theorem for PolynomialsPolynomialsLet 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.


EXAMPLE: Approximating a Real Zeroa. 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

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

Solution

This sign change shows that the polynomial function has a real zero between 2 and 3.

and

f (3) 33 2 3 5 27 6 5 16.f (3) is positive. f (3) is positive.

f (2) 23 2 2 5 8 4 5 1f (2) is negative. f (2) is negative.


EXAMPLE: Approximating a Real Zero

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.

Solution

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

The sign change indicates that f has a real zero between 2 and 2.1.

x f (x) x3 2x 5

2 f (2) 23 2(2) 5 1

2.1 f (2.1) (2.1)3 2(2.1) 5 0.061Sign changeSign change

Sign changeSign change


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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

EXAMPLE: Approximating a Real Zero

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 sub-intervals. Then we evaluate f at each endpoint and look for a sign change.

Solution

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

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.

Sign changeSign change



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 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.

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.

The Fundamental Theorem of AlgebraThe Fundamental Theorem of Algebra

The Linear Factorization TheoremThe Linear Factorization Theorem

The Linear Factorization Theorem The Linear Factorization Theorem If f (x) anx

n an1xn1 … 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.

The Linear Factorization Theorem The Linear Factorization Theorem If f (x) anx

n an1xn1 … 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.

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.



EXAMPLE: Finding a Polynomial Function with Given Zeros

Find a fourth-degree polynomial function f (x) with real coefficients that has2, 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.

an(x 2)(x 2)(x i)(x i) Use the given zeros: c1 2, c2 2, c3 i, and, from above, c4 i.

f (x) an(x c1)(x c2)(x c3)(x c4) This is the linear factorization for a fourth-degree polynomial.

an(x2 4)(x2 i) Multiply

f (x) an(x4 3x2 4) Complete the multiplication

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EXAMPLE: Finding a Polynomial Function with Given Zeros

Find a fourth-degree polynomial function f (x) with real coefficients that has2, and i as zeros and such that f (3) 150.

Substituting 3 for an in the formula for f (x), we obtain

f (x) 3(x4 3x2 4). Equivalently,

f (x) 3x4 9x2 12.

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

4.5: more on zeros of polynomial functions the upper and lower bound theorem helps us rule out many...

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