Chapter 3: Polynomial Functions 3.1 Complex Numbers 3.2 Quadratic Functions and Graphs 3.3 Quadratic Equations and Inequalities 3.4 Further Applications of Quadratic Functions and Models 3.5 Higher-Degree Polynomial Functions and Graphs 3.6 Topics in the Theory of Polynomial Functions (I) 3.7 Topics in the Theory of Polynomial Functions (II)

Topics in the Theory of Polynomial Functions (I) The Intermediate Value Theorem If P(x) defines a polynomial function with only real coefficients, and if, for real numbers a and b, the values P(a) and P(b) are opposite in sign, then there exists at least one real zero between a and b.

Applying the Intermediate Value Theorem Example Show that the polynomial function defined by P(x) = x3 − 2x2 − x + 1 has a real zero between 2 and 3.

Applying the Intermediate Value Theorem Example Show that the polynomial function defined by P(x) = x3 − 2x2 − x + 1 has a real zero between 2 and 3. Analytic Solution Evaluate P(2) and P(3). P(2) = 23 − 2(2)2 − 2 + 1 = −1 P(3) = 33 − 2(3)2 − 3 + 1 = 7 Since P(2) = −1 and P(3) = 7 differ in sign, the intermediate value theorem assures us that there is a real zero between 2 and 3.

Applying the Intermediate Value Theorem We see that the zero lies between 2.246 and 2.247 since there is a sign change in the function values. Caution If P(a) and P(b) do not differ in sign, it does NOT imply that there is no zero between a and b.

Division of Polynomials Example Divide the polynomial 3x3 − 2x + 5 by x − 3.

Division of Polynomials Example Divide the polynomial 3x3 − 2x + 5 by x − 3.

Division of Polynomials

Division of Polynomials

Synthetic Division Example Divide the polynomial 3x3 − 2x + 5 by x − 3.

Synthetic Division This abbreviated form of long division is called synthetic division.

Using Synthetic Division Example Use synthetic division to divide 5x3 − 6x2 − 28x + 8 by x + 2. nklon

Using Synthetic Division

The Remainder Theorem Remainder Theorem If a polynomial P(x) is divided by x − k, the remainder is equal to P(k). Example Use the remainder theorem and synthetic division to find P(−2) if P(k) = − x4 + 3x2 − 4x − 5. Note: Can be solved by substitution and or by synthetic division

k is Zero of a Polynomial Function if P(k) = 0 Example Decide whether the given number is a zero of P.

k is Zero of a Polynomial Function if P(k) = 0 Example Decide whether the given number is a zero of P.

k is Zero of a Polynomial Function if P(k) = 0 Graphical Solution Y1 = P(x) in part (a) Y1(2) = P(2) in part (a) Y2 = P(x) in part (b) Y2 (−2) = P(−2) in part(b)

The Factor Theorem The Factor Theorem The polynomial P(x) has a factor x − k if and only if P(k) = 0.

Example using the Factor Theorem Determine whether the second polynomial is a factor of the first. P(x) = 4x3 + 24x2 + 48x + 32; x + 2

Example using the Factor Theorem Determine whether the second polynomial is a factor of the first. P(x) = 4x3 + 24x2 + 48x + 32; x + 2 Solution Use synthetic division with k = −2.

Relationships Among x-Intercepts, Zeros, and Solutions Graph P in a suitable viewing window and locate the x-intercepts. Solve the polynomial equation 2x3 + 5x2 − x − 6 = 0.

Relationships Among x-Intercepts, Zeros, and Solutions

Relationships Among x-Intercepts, Zeros, and Solutions The calculator will determine the x-intercepts: −2, −1.5, and 1.

Topics in the Theory of Polynomial Functions Example Find a polynomial having zeros 3 and 2 + i that satisfies the requirement P(−2) = 4.

Topics in the Theory of Polynomial Functions Example Find a polynomial having zeros 3 and 2 + i that satisfies the requirement P(−2) = 4. Solution Since 2 + i is a zero, so is 2 − i. A general solution is

Zeros of a Polynomial Function Number of Zeros Theorem A function defined by a polynomial of degree n has at most n distinct (unique) complex zeros. Example Find all complex zeros of P(x) = x4 − 7x3 + 18x2 − 22x + 12 given that 1 − i is a zero.

Zeros of a Polynomial Function Number of Zeros Theorem A function defined by a polynomial of degree n has at most n distinct (unique) complex zeros. Example Find all complex zeros of P(x) = x4 − 7x3 + 18x2 − 22x + 12 given that 1 − i is a zero.

Zeros of a Polynomial Function Using the Conjugate Zeros Theorem, 1 + i is also a zero.

Multiplicity of a Zero The multiplicity of the zero refers to the number of times a zero appears.

Polynomial Function Satisfying Given Conditions Example Find a polynomial function with real coefficients of lowest possible degree having a zero 2 of multiplicity 3, a zero 0 of multiplicity 2, and a zero i of single multiplicity.

Polynomial Function Satisfying Given Conditions Example Find a polynomial function with real coefficients of lowest possible degree having a zero 2 of multiplicity 3, a zero 0 of multiplicity 2, and a zero i of single multiplicity. Solution By the conjugate zeros theorem, −i is also a zero.

Rough Sketch using Multiplicities and End Behavior Draw a rough sketch of the polynomial The following figure illustrates some conclusions. P(x) = (x + 3)2(x − 2)2 b. P(x) = (x + 3)2(x − 2)3 c. P(x) = -x2(x − 1)(x + 2)2

Sketching a Graph of a Polynomial Function by Hand Example Sketch P(x) = −2(x + 4)2(x + 3)(x − 1)2 by hand.

Sketching a Graph of a Polynomial Function by Hand Example Sketch P(x) = −2(x + 4)2(x + 3)(x − 1)2 by hand. Solution The dominating term is −2x5, so the end behavior will rise on the left and fall on the right. Because −4 and 1 are x-intercepts determined by zeros of even multiplicity, the graph will be tangent to the x-axis at these x-intercepts. The y-intercept is −96.

The Rational Zeros Theorem Example P(x) = 6x4 + 7x3 − 12x2 − 3x + 2 List all possible rational zeros. Use a graph to eliminate some of the possible zeros listed in part (a). Find all rational zeros and factor P(x).

The Rational Zeros Theorem From the graph, the zeros are no less than −2 and no greater than 1. Also, −1 is clearly not a zero since the graph does not intersect the x-axis at the point (−1, 0).

The Rational Zeros Theorem Show that 1 and −2 are zeros.

Descartes’ Rule of Sign Descartes’ Rule of Signs Let P(x) define a polynomial function with real coefficients and a nonzero constant term, with terms in descending powers of x. The number of positive real zeros either equals the number of variations in sign occurring in the coefficients of P(x) or is less than the number of variations by a positive even integer. The number of negative real zeros either equals the number of variations in sign occurring in the coefficients of P(x) or is less than the number of variations by a positive even integer.

Applying Descartes’ Rule of Signs Example Determine the possible number of positive real zeros and negative real zeros of P(x) = x4 − 6x3 + 8x2 + 2x − 1. We first consider the possible number of positive zeros by observing that P(x) has three variations in signs.

Boundedness Theorem Boundedness Theorem Let P(x) define a polynomial function of degree n  1 with real coefficients and with a positive leading coefficient. If P(x) is divided synthetically by x − c, and if c > 0 and all numbers in the bottom row of the synthetic division are nonnegative, then P(x) has no zero greater than c; if c < 0 and the numbers in the bottom row of the synthetic division alternate in sign (with 0 considered positive or negative, as needed), then P(x) has no zero less than c.

Using the Boundedness Theorem Example Show that the real zeros of P(x) = 2x4 – 5x3 + 3x + 1 satisfy the following conditions. a) No real zero is greater than 3. b) No real zero is less than –1.