1. Upper and Lower Bounds for Roots
3.5: 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.

2. EXAMPLE: Finding Bounds for the Roots
3.5: 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 13 35
All numbers in this row are nonnegative.
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3. EXAMPLE: Finding Bounds for the Roots
3.5: 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 9
-24 42 -9 26 13 35
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.)

4. The Intermediate Value Theorem
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.

5. EXAMPLE: Approximating a Real Zero
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
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) =  = = 16.
f (3) is positive.
f (2) =  = = -1
f (2) is negative.

6. EXAMPLE: Approximating a Real Zero
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) = (2) = -1
2.1
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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7. EXAMPLE: Approximating a Real Zero
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 sub- intervals. Then we evaluate f at each endpoint and look for a sign change.
f (2.00) = -1
f (2.04) =
f (2.08) =
f (2.01) =
f (2.05) =
f (2.09) =
f (2.02) =
f (2.06) =
f (2.1) = 0.061
f (2.03) =
f (2.07) =
Sign change
The sign change indicates that f has a real zero between 2.09 and Correct to the nearest tenth, the zero is 2.1.

8. The Fundamental Theorem of Algebra
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.

9. The Linear Factorization Theorem
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.

10. EXAMPLE: Finding a Polynomial Function with Given Zeros
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.
= 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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11. EXAMPLE: Finding a Polynomial Function with Given Zeros
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(  ) = To find an, use the fact that f (3) = -150.
an( ) = 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 + 9x