Solving Polynomial Functions
Factoring a Polynomial
• Remember Three Types of Problems
– Difference/Sum of Two Cubes
– Grouping
– Quadratic Like
Practice Problems
•
1.
2.
3.
Factor and solve
x3 – 64
x3 + 6x2 -4x -24=0
4w4 + 40w2- 44=0
Using Division to Find a Zero
• If you know one zero, you can use division
to find another.
• Remainder Theorem- If a polynomial f(x) is
divided by (x-k) then the remainder is
r=f(x)
• Two forms of polynomial division: long
division and synthetic division
Long Division- can be used on any
polynomial
• Divide f(x)=3x4 – 5x3 + 4x – 6
by (x2 -3x + 5)
x2 -3x + 5 3x4 – 5x3 + 0x2 + 4x – 6
Practice
• Divide f(x)=x3 + 5x2 – 7x + 2 by x-2
Synthetic Division
• Can be used to divide any polynomial by a
divisor of the form x – k
• To set up synthetic division, list the
coefficients in a row.
F(x)= 2x3 + x2 -8x + 5 by x + 3
-3
2
1
-8
5
Factor Theorem- a polynomial f(x)
has a factor x-k if f(k)=0
Factor the polynomial
F(x) = 3x3 – 4x2 – 28x -16 completely
given that x+ 2 if a factor.
The profit P ( in millions of dollars) for a shoe manufacturer can be
modeled by P= -21x3 + 46x where x is the number of shoes
produced (in millions). The company produces 1 million shoes and
makes a profit of 25,000,000 but would like to cut back on
production. What lesser number of shoes could the company make
and still make the same profit?
Rational Zero Theorem
• One way to narrow down the possible
zeros of a function is to use the Rational
Zero Theorem.
• If f(x)= anxn + …anx + a0 has integer
coefficients, then every rational zero of f
has the following form
p = factors of constant term a0
q factors of leading coefficient an
List the possible rational zeros of f using
rational zero theorem.
F(x) = x3 + 2x2 – 11x - 12
Find all real zeros of f(x)= x3 – 8x2+ 11x + 20
Find all real zeros of
f(x)= 10x4 - 11x3 – 42x2+ 7x + 12
Step 1: List possible rational zeros.
Step 2: Use graphing calculator to narrow
down choice\
Step 3: Use synthetic division to test zero
10x4 - 11x3 – 42x2+ 7x + 12
10
-11
-42
7
12
Fundamental Theorem of Algebra
Theorem: If f(x) is a polynomial of degree n
where n >0 then the equation f(x)=0 has at least
one solution in the set of complex numbers.
Corollary: if f(x) is a polynomial of degree n where
n>0 then 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.
Complex Conjugate Theorem
• If f is a polynomial function with real
coefficients and a + bi is an imaginary zero
of f, then a- bi is a zero of f.
Irrational Conjugate Theorem
Suppose f is a polynomial function with
rational coefficients and a and b are
rational numbers such that √ b is irrational.
If a + √b is a zero of f , then a – √b is also
a zero of f.
Find the zeros
F(x) = x5 – 4x4 + 4x3 + 10x2 – 13x - 14
Write a polynomial function f of least degree
that has rational coefficients , a leading
coefficient of 1 and 3 and 2 + √5 as zeros.
Set up factors
F(x)= (x – 3)(x – (2 + √5 ))(x – (2 - √5 )
Write the polynomial function f of least degree that has
rational coefficients a leading coefficient of 1 and the given
zeros
1. -1, 2, 4
2. 3, 3-i
Descartes Rule of Signs
Let f(x) = anxn + …anx + a0 be a polynomial
function with real coefficients.
-The number of positive real zeros of f is equal to
the number of changes in sign of 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 coefficients of
f(-x) or is less than this by an even number.
Determine the possible numbers of positive
real zeros, negative real and imaginary
zeros for the function
F(x)= x6 – 2x5 + 3x4 – 10x3 – 6x2 -8x -8
Pos. real zeros
Neg, real zeros
Imag. zeros
Total zeros