Transcript 7.5 Roots and Zeros

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=4
x²-5x+4=0 (x-4)(x-1)=0 Roots: 1,4
x²+2x+8=0 x=-1i√3 Roots are imaginary
x³+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 + 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)= -x5 – 3x4 + x2 + x + 6
There 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 0
Negative 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 Zeros
Use Descarteś Rule of Signs to know what
to test.
Example: f(x) = x3 – 6x2 +10x – 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 1st one, it gets easier from there. You
can use synthetic substitution to determine
if they are roots.
Continued
• f(x) = x3 – 6x2 +10x – 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
-3
1
-5
4
1
-2
2
0
Use remaining polynomial to find others. x²-2x+2=0
(2)  4  4(1)(2) 2  2i
x

 1 i
2
2
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
Homework
p. 375
14-32 even