Transcript Academy Algebra II 5.7: Apply the Fundamental Theorem of Algebra

Do Now: Find all real zeros of the function.
f ( x)  2 x  5 x  18 x  19 x  42
4
3
2
Academy Algebra II
5.7: Apply the Fundamental
Theorem of Algebra
HW Friday: p.384 (12-18 even)
Test 5.5-5.9: Tuesday, 12/2
Descartes’ Rule of Signs
• The number of positive real zeros of f is
equal to the number of sign changes in
the sign of the coefficients of f(x) or is
less than this by an even number.
• The number of negative real zeros of f is
equal to the number of changes in sign of
the coefficients of f(-x) or is less than this
by an even number.
Fundamental Theorem of Algebra
• If f(x) is a polynomial with degree of
n (where n>0), then the equation
f(x) = 0 has at least one solution.
• Corollary: The equation f(x) = 0 has
exactly n solutions provided each
solution repeated twice is counted as
2 solutions, each solution repeated
three times is counted as 3 solutions,
and so on.
Fundamental Theorem of Algebra
• Example: x3 – 5x2 – 8x + 48 = 0
(x+3)(x – 4)2= 0
x = -3 and x = 4
This equation only has two distinct
solutions: -3 and 4.
Because the factor (x – 4) appears twice,
you can count the solution 4 twice. With
the repeated solution the third-degree
equation has three solutions.
Solve the equation.
x3 + 5x2 + 4x + 20 = 0
Find the zeros of the
polynomial function.
f(x) = x4 – 8x3 + 18x2 – 27
Find all zeros of the
polynomial function.
f(x) = x5 – 2x4 + 8x2 – 13x + 6
Conjugates Theorem
• If a  bi is an imaginary zero of
function f, then a  bi is also a zero.
• If a  b is a zero of function f where
is irrational, then a  b is also a
zero.
b
Write a polynomial function f of least
degree that has rational coefficients, a
leading coefficient of 1, & the given zeros.
1.) -1, 2, 4
2.) 4, 1 5
Write a polynomial function f of least
degree that has rational coefficients, a
leading coefficient of 1, & the given zeros.
3.) 2, 2i, 4  6