3.3 Zeros of Polynomials

Complete factorization theorem for Polynomials If f(x) is a polynomial of degree n > 0, then there exist n complex numbers c1, c2, …., cn such that f (x) = a (x – c1) (x – c2)…(x - cn)

Ex. Find a polynomial of degree 3 that has the indicated zeros and satisfies the given condition. Zeros at x = -5, 2, 4 f(3) = -24

Solution f (x) = a (x + 5) (x-2) (x-4) Solve for a by plugging in f(3) = -24 -24 = (3+5)(3-2)(3-4) -24 = a8(1)(-1) 3 = a f(x) = 3 (x+5)(x-2)(x-4) *multiply out f (x) = 3x3 – 3x2 – 66x + 120

3.3 The Fundamental Theorem of Algebra Every function defined by a polynomial of degree 1 or more has at least one complex (real or imaginary) zero. Conjugate Zeros Theorem If P(x) is a polynomial having only real coefficients, and if a + bi is a zero of P, then the conjugate a – bi is also a zero of P.

3.3 Multiplicity of a Zero The multiplicity of the zero refers to the number of times a zero appears. e.g. – x = 0 leads to a single zero – (x + 2)2 leads to a zero of –2 with multiplicity two – (x – 1)3 leads to a zero of 1 with multiplicity three

3.3 Polynomial Function Satisfying Given Conditions Example Find a polynomial function with real coefficients of lowest possible degree having a leading coefficient 1, 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. This is one of many such functions. Multiplying P(x) by any nonzero number will yield another function satisfying these conditions.

3.3 Observation: Parity of Multiplicities of Zeros Observe the behavior around the zeros of the polynomials The following figure illustrates some conclusions. By observing the dominating term and noting the parity of multiplicities of zeros of a polynomial in factored form, we can draw a rough graph of a polynomial by hand.

3.3 Sketching a Polynomial Function by Hand Example 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.

Descartes’ Rule of Sign Let P(x) define a polynomial function with real coefficients and a nonzero constant term, with terms in descending powers of x. (a) The number of positive real zeros of P 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. (b) The number of negative real zeros of P is either 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.

3.3 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. + x4 – 6x3 + 8x2 + 2x – 1 Thus, by Descartes’ rule of signs, f has either 3 or 3 – 2 = 1 positive real zeros. For negative zeros, consider the variations in signs for P(x). P(x) = (x)4 – 6(x)3 + 8(x)2 + 2(x)  1 = x4 + 6x3 + 8x2 – 2x – 1 Since there is only one variation in sign, P(x) has only one negative real root. 1 2 3

3.3 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 (a) 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; (b) 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.

3.3 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. Solution a) c > 0 b) c < 0 All are nonegative. No real zero greater than 3. The numbers alternate in sign. No zero less than 1.