7 5 Roots and Zeros Objectives 1 Determine

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 + … + a 1 x + a 0 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 x 2 – 7 x + 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=4 x²-5 x+4=0 (x-4)(x-1)=0 Roots: 1, 4 x²+2 x+8=0 x=-1 i√ 3 Roots are imaginary x³+x²-12 x=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) = x 5 – 3 x 4 + x 2 – x + 6 How many sign changes in coefficients? 4 So there are 4, 2, or 0 positive real roots. Now find P(-x) (for odd exponents, change the sign) P(-x)= -x 5 – 3 x 4 + x 2 + x + 6 There is 1 sign change. There 1 negative 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+4 x³-2 x²-x-1 Positive Real: 2 or 0 Negative Real: p(-x)= -x -4 x³-2 x²+x-1 2 or 0 Imaginary: 6, 4, or 2

Try one f(x) = x 3 – 6 x 2 +10 x – 8

Finding Zeros Use Descarteś Rule of Signs to know what to test. Example: f(x) = x 3 – 6 x 2 +10 x – 8 Roots: 3 or 1 positive real, 0 negative real So there are 2 or 0 imaginary roots. Choose some positive numbers to sub. in to determine if they are roots. Once you find the 1 st one, it gets easier from there. You can use synthetic substitution to determine if they are roots.

Continued • f(x) = x 3 – 6 x 2 +10 x – 8 Use a table to organize synthetic substitution. x 1 -6 10 -8 1 1 -5 5 -3 4 is a root!! 2 1 -4 2 -4 3 1 -5 4 1 -2 2 0 Use remaining polynomial to find others. x²-2 x+2=0

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) = x 4 – 9 x 3 + 24 x 2 – 6 x – 40

Homework p. 375 14 -32 even