Transcript The Fundamental Theorem of Algebra

Today we will be finding zeros….
So what is a zero….
 http://www.youtube.com/watch?v=hXE2JNf_fd8
The Fundamental
Theorem of Algebra
5.7
IF F(X) IS A POLYNOMIAL
OF DEGREE N WHERE N>0,
THEN THE EQUATION
F(X) = O HAS AT LEAST
ONE SOLUTION IN THE SET
OF COMPLEX NUMBERS.
Some polynomials have repeated
solutions.
For example: x – 6x + 9 = 0
(x – 3) (x – 3) = 0
x=3
2
Since the polynomial is a
quadratic there must be 2
zeros, 3 is a double root.
How many solutions does each
equation have?
5
x
–
2
7x +
9=0
x4 + 9x = 18
Complex Conjugates Theorem
If f (x) is a polynomial
function with real
coefficients, and a+bi is an
imaginary zero of f(x), then
a-bi is also a zero of f(x).
(imaginary numbers always come in pairs)
Irrational Conjugates Theorem
Suppose f(x) is a polynomial
function with rational
coefficients, and a and b are
rational numbers such that b.5
is irrational. If a + b.5 is a zero
of f(x), then a - b.5 is also a
zero of f(x).
Some possibilities for zeros (roots)
Equation
Real Zeros
Imaginary Zeros
Linear
1
0
Quadratic
2
0
0
2
3
0
1
2
4
0
2
2
0
4
5
0
3
2
1
4
Cubic
Quartic
Quintic
Find the zeros.
f(x) = x 3 – x2 + 4x – 4 = 0
g(x) = x 4 + 2x 3 – 5x 2 – 4x + 6 = 0
Write a polynomial function of least
degree with a leading coefficient of 1.
2, 3, -4
(x – 2)(x – 3)(x + 4) = 0
2
(x – 5x + 6)(x + 4) = 0
3
2
x - x – 14x + 24 = 0
Write a polynomial function of least
degree with a leading coefficient of 1.
7, i
(x – 7)(x – i)(x + i) = 0
2
(x – 7)(x + 1) = 0
x3–7x2 +x–7=0
Descartes’ Rule of Signs
Let f(x) be a polynomial function
with real coefficients.
 The number of positive real zeros of f(x)
is equal to the number of sign changes
of the coefficients of f(x) or is less than
this by an even number.
 The number of negative real zeros of
f (x) is equal to the number of sign
changes of the coefficients of f(-x) or is
less than this by an even number.
huh
???
Using Descartes’ Rule of Signs….
f(x) = x6-2x 5+3x 4 – 10x 3-6x 2 -8x -8
3 sign changes, so f has 3 or 1 positive real roots
f(-x) = x6+2x 5+3x 4 + 10x 3-6x 2 +8x -8
3 sign changes, so f has 3 or 1 negative real
roots
g(x) = -x5 – 5x4 + 7x3 – 4x2 – 8x + 9
 Determine the possible number of
positive real zeros, negative real zeros,
and imaginary zeros of the function
given above.
 g(x) = -x5 – 5x4 + 7x3 – 4x2 – 8x + 9
 g(-x) = x5 – 5x4 - 7x3 – 4x2 + 8x + 9
Factoring polynomials on
youtube
HTTP://WWW.YOUTUBE.COM/WATCH?V=X9V
QYKS7RB0