7.5 Roots and Zeros

Objectives:1. Determine the number and

type of roots for a polynomial function.

2. Find the zeros of a polynomial function

Zeros, Factors, and Roots

• Let f(x) = anxn + … + a1x + a0 be a polynomial function. Then– c is a zero of the polynomial function f(x),– x – c is a factor of the polynomial f(x), and – c is a root or solution of the polynomial

function f(x) = 0.• In addition, if c is a real number, then (c,0)

is an x-intercept of the graph of f(x).

Examples• Given x2 – 7x + 12 .• It factors as (x – 4)(x – 3)• If we are solving for the zeros, then

(x – 4)(x – 3) = 0 or x = 4 and x = 3. These are zeros of the function

• Given x = 4, then x – 4 and x = 3, then x – 3 is a factor.

• x = 3 and x = 4 are roots or solutions of the function.

• The graph crosses the x-axis at (3,0) and (4,0)

Roots

• When you solve a polynomial equation with degree greater than zero, it may have– one or more real roots, or– no real roots (the roots are imaginary).

Fundamental Theorem of Algebra

Every polynomial equation with degree greater than zero has at least one root in the set of complex numbers.

Examples:x-4=0 Root: x=4x²-5x+4=0 (x-4)(x-1)=0 Roots: 1,4x²+2x+8=0 x=-1i√3 Roots are imaginaryx³+x²-12x=0 x(x+4)(x-3)=0 Roots: 0, -4, 3

Corollary to the Fundamental Theorem of Algebra

A polynomial equation of the form P(x) = 0 of degree n with complex coefficients has exactly n roots in the set of complex numbers.

Degree=roots

Descarteś Rule of Signs

• If P(x) is a polynomial with real coefficients whose terms are arranged in descending powers of the variable,– the number of positive real zeros of y = P(x) is the

same as the number changes in sign of the coefficients of the terms, or is less than this by an even number, and

– the number of negative real zeros of y = P(x) is the same as the number of changes in sign of the coefficients of the terms of P(-x), or is less than this number by an even number.

Find the number of Positive and Negative Roots

P(x) = x5 – 3x4 + x2 – x + 6How many sign changes in coefficients? 4So there are 4, 2, or 0 positive real roots.Now find P(-x) (for odd exponents, change

the sign)P(-x)= -x5 – 3x4 + x2 + x + 6There is 1 sign change. There 1negative real

root. (can’t be less because it can’t be reduced by a multiple of 2)

Example

State the possible number of positive real zeros, negative real zeros, and imaginary zeros of p(x)=-x⁶+4x³-2x²-x-1

Positive Real: 2 or 0Negative Real: p(-x)= -x -4x³-2x²+x-1

2 or 0 Imaginary: 6, 4, or 2

Try one

f(x) = x3 – 6x2 +10x – 8

Finding ZerosUse Descarteś Rule of Signs to know what

to test.Example: f(x) = x3 – 6x2 +10x – 8Roots: 3 or 1 positive real, 0 negative realSo there are 2 or 0 imaginary roots.Choose some positive numbers to sub. in to

determine if they are roots. Once you find the 1st one, it gets easier from there. You can use synthetic substitution to determine if they are roots.

Continued

x 1 -6 10 -81 1 -5 5 -32 1 -4 2 -43 1 -3 1 -54 1 -2 2 0

• f(x) = x3 – 6x2 +10x – 8Use a table to organize synthetic substitution.

4 is a root!!

Use remaining polynomial to find others. x²-2x+2=0

( 2) 4 4(1)(2) 2 2 12 2

ix i

Complex Conjugates Theorem

• Complex Roots always come in PAIRS• The complex conjugate theorem states

Suppose a and b are real numbers with b ≠ 0. If a + bi is a zero of a polynomial function with real coefficients, then a – bi is also a zero of the function.

Try one

• Find the zeros of f(x) = x4 – 9x3 + 24x2 – 6x – 40

Homeworkp. 375

14-32 even