POLYNOMIAL EQUATIONS
(Part 1)
The solutions to a polynomial equation, f(x) = 0, are
the zeros of the corresponding function, y = f(x).
POLYNOMIAL FUNCTION:
POLYNOMIAL EQUATION:
f(x) = anxn + an-1xn-1 + ... + a1x + a0
anxn + an-1xn-1 + ... + a1x + a0 = 0
f(x) = 0
factorable
factor by
grouping
unfactorable
find a factor using
the factor theorem
factor the quotient
use technology
(graphing calculator)
use the quadratic
formula for the quotient
IF A POLYNOMIAL EQUATION IS NOT FACTORABLE, IT MAY STILL HAVE A SOLUTION!!
Possible factors of a polynomial function,
f(x) = anxn + an-1xn-1 + ... + a1x + a0,
are in the form p where p is a factor of a0
q
and q is a factor of an.
Ex.
3x3 + 2x2 + 17x – 6 = 0
Ex 
State the zeros of the following functions:
a)
Ex 
f(x) = x(x + 3)(2x – 4)
Solve each polynomial equation:
a)
x3 + 2x2 – x – 2 = 0
b)
x3 + 4x2 + x – 6 = 0
c)
x3 – 2x2 + 6x + 11 = x2 + 12x – 5
b)
f(x) = (x2 – 9)(x2 + 25)
d)
x4 + 5x2 + 1 = 0
e)
x3 – 2x + 3 = 0
Homework:
p.204–206 2ac, 6, 7abf, 8ac, 9bd, 15, 16
Solve each of the following:
1.
2.
3.
0 = x3 + 3x2 – 7x – 6
0 = x3 – 4x2 – 2x – 15
0 = 6x3 – 13x2 + 7x – 1
Answers:
1.
x = 2,
2.
x=5
3.
x=
 5  13
2
1 5  13
,
6
2