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```2.5 Zeros of Polynomial Functions
Roots & Zeros of Polynomials
How the roots, solutions,
zeros, x-intercepts and
factors of a polynomial
function are related.
Polynomials
A Polynomial Expression can be a
monomial or a sum of monomials. The
Polynomial Expressions that we are
discussing today are in terms of one
variable.
In a Polynomial Equation, two
polynomials are set equal to each other.
Factoring Polynomials
Terms are Factors of a Polynomial if, when
they are multiplied, they equal that
polynomial:
x  2 x  15  ( x  3)( x  5)
2

(x - 3) and (x + 5) are Factors
of the polynomial
x  2 x  15
2
Since Factors are a Product...
…and the only way a product can equal zero
is if one or more of the factors are zero…
…then the only way the polynomial can
equal zero is if one or more of the factors are
zero.
Solving a Polynomial Equation
Rearrange the terms to have zero on one side:
x  2 x  15  x  2 x  15  0
2
Factor:
2
( x  5)( x  3)  0
Set each factor equal to zero and solve:
( x  5)  0 and
x  5
( x  3)  0
x 3
The only way that x2 +2x - 15 can = 0 is if x = -5 or x = 3
Solutions/Roots a Polynomial
Setting the Factors of a Polynomial
Expression equal to zero gives the
Solutions to the Equation when the
polynomial expression equals zero.
Another name for the Solutions of a
Polynomial is the Roots of a
Polynomial !
Zeros of a Polynomial Function
A Polynomial Function is usually written in
function notation or in terms of x and y.
2
f (x)  x  2x  15
or
2
y  x  2x  15
The Zeros of a Polynomial Function are the
solutions to the equation you get when you
set the polynomial equal to zero.
Zeros of a Polynomial Function
The Zeros of a Polynomial
Function ARE the Solutions to
the Polynomial Equation when
the polynomial equals zero.
Graph of a Polynomial
Function
Here is the graph of our
polynomial function:
2
y  x  2x  15
The Zeros of the Polynomial are the values of x
when the polynomial equals zero. In other words,
the Zeros are the x-values where y equals zero.
x-Intercepts of a Polynomial
The points where y = 0 are
called the x-intercepts of the
graph.
The x-intercepts for our graph
are the points...
(-5, 0)
and
y  x 2  2x  15
(3, 0)
x-Intercepts of a Polynomial
When the Factors of a Polynomial
Expression are set equal to zero, we get
the Solutions or Roots of the Polynomial
Equation.
The Solutions/Roots of the Polynomial
Equation are the x-coordinates for the
x-Intercepts of the Polynomial Graph!
Factors, Roots, Zeros
For our Polynomial Function:
2
y  x  2x  15
The Factors are:
(x + 5) & (x - 3)
The Roots/Solutions are:
x = -5 and 3
The Zeros are at:
(-5, 0) and (3, 0)
Roots & Zeros of Polynomials II
Finding the Roots/Zeros of Polynomials:
• The Fundamental Theorem of Algebra
•
Descartes’ Rule of Signs
• The Complex Conjugate Theorem
Fundamental Theorem Of Algebra
Every Polynomial Equation with a degree
higher than zero has at least one root in the
set of Complex Numbers.
COROLLARY:
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.
Real/Imaginary Roots
If a polynomial has ‘n’ complex roots will its
graph have ‘n’ x-intercepts?
3
y  x  4x
In this example, the
degree n = 3, and if we
factor the polynomial, the
roots are x = -2, 0, 2. We
can also see from the
graph that there are 3
x-intercepts.
Real/Imaginary Roots
Just because a polynomial has ‘n’ complex
roots doesn’t mean that they are all Real!
In this example,
however, the degree is
still n = 3, but there is
only one Real x-intercept
or root at x = -1, the
other 2 roots must have
imaginary components.
y  x 3  2x 2  x  4
Descartes’ Rule of Signs
Arrange the terms of the polynomial P(x) in
descending degree:
• The number of times the coefficients of the terms
of P(x) change sign = the number of Positive Real
Roots (or less by any even number)
• The number of times the coefficients of the terms
of P(-x) change sign = the number of Negative
Real Roots (or less by any even number)
In the examples that follow, use Descartes’ Rule of Signs to
predict the number of + and - Real Roots!
Find Roots/Zeros of a
Polynomial
We can find the Roots or Zeros of a polynomial by
setting the polynomial equal to 0 and factoring.
Some are easier to
factor than others!
3
f (x)  x  4x
2
 x(x  4)
 x(x  2)(x  2)
The roots are: 0, -2, 2
Find Roots/Zeros of a
Polynomial
If we cannot factor the polynomial, but know one of the
roots, we can divide that factor into the polynomial. The
resulting polynomial has a lower degree and might be
easier to factor or solve with the quadratic formula.
f (x)  x  5x  2x 10
3
2
one root is x  5
(x - 5) is a factor
We can solve the resulting
polynomial to get the other 2 roots:
x  2,  2
2
x 2
3
2
x  5 x  5x  2x  10

x 3  5x 2
 2x  10
2x  10
0
Complex Conjugates Theorem
Roots/Zeros that are not Real are Complex with an
Imaginary component. Complex roots with
Imaginary components always exist in Conjugate
Pairs.
If a + bi (b ≠ 0) is a zero of a polynomial function,
then its Conjugate, a - bi, is also a zero of the
function.
Find Roots/Zeros of a
Polynomial
If the known root is imaginary, we can use the Complex
Conjugates Theorem.
Ex: Find all the roots of
If one root is 4 - i.
f (x)  x 3  5x 2  7x  51
Because of the Complex Conjugate Theorem, we know that
another root must be 4 + i.
Can the third root also be imaginary?
Consider…
Descartes: # of Pos. Real Roots = 2 or 0
Descartes: # of Neg. Real Roots = 1
Example (con’t)
3
2
Ex: Find all the roots of f (x)  x  5x  7x  51
If one root is 4 - i.
If one root is 4 - i, then one factor is [x - (4 - i)], and
Another root is 4 + i, & another factor is [x - (4 + i)].
Multiply these factors:
 x   4  i    x   4  i    x 2  x  4  i   x  4  i    4  i  4  i 
 x 2  4 x  xi  4 x  xi  16  i 2
 x 2  8 x  16  (1)
 x 2  8 x  17
Example (con’t)
3
2
Ex: Find all the roots of f (x)  x  5x  7x  51
If one root is 4 - i.
2
If the product of the two non-real factors is x  8x 17
then the third factor (that gives us the neg. real root) is
2
the quotient of P(x) divided by x  8x 17 :
x 3
x 2  8x  17 x 3  5x 2  7x  51

x 3  5x 2  7x  51
0
The third root
is x = -3
Finding Roots/Zeros of
Polynomials
We use the Fundamental Thm. Of Algebra, Descartes’ Rule
of Signs and the Complex Conjugate Thm. to predict the
nature of the roots of a polynomial.
We use skills such as factoring, polynomial division and the
quadratic formula to find the zeros/roots of polynomials.
In future lessons you will learn
other rules and theorems to predict
the values of roots so you can solve
higher degree polynomials!
The Rational Zero Theorem
The Rational Zero Theorem gives a list of possible rational zeros of a
polynomial function. Equivalently, the theorem gives all possible rational roots
of a polynomial equation. Not every number in the list will be a zero of the
function, but every rational zero of the polynomial function will appear
somewhere in the list.
The Rational Zero Theorem
p
q
p
n
n-1
If f(x) 
a
x

a
x
… a1x  a0 has integer coefficients and
n
n-1
q
(where is reduced) is a rational zero, then p is a factor of the constant
term a0 and q is a factor of the leading coefficient an.
EXAMPLE: Using the Rational Zero Theorem
List all possible rational zeros of f(x)  15x3  14x2  3x – 2.
Solution
The constant term is –2 and the leading coefficient is 15.
Factors of the constant term,  2
Factors of the leading coefficient, 15
1,  2

1,  3,  5,  15
Possible rational zeros 
 1,  2,
 13 ,  23 ,
Divide 1
and 2
by 1.
Divide 1
and 2
by 3.
 15 ,  52 ,
Divide 1
and 2
by 5.
1 ,  2
 15
15
Divide 1
and 2
by 15.
There are 16 possible rational zeros. The actual solution set to f(x)  15x3 
14x2  3x – 2 = 0 is {-1, 1/3, 2/5}, which contains 3 of the 16 possible solutions.
EXAMPLE:
Solve:
Solving a Polynomial Equation
x4  6x2  8x + 24  0.
Solution Because we are given an equation, we will use the word "roots,"
rather than "zeros," in the solution process. We begin by listing all possible
rational roots.
Factors of the constant term, 24
Factors of the leading coefficient, 1
1,  2  3,  4,  6,  8,  12,  24

1
 1,  2  3,  4,  6,  8,  12,  24
Possible rational zeros 
EXAMPLE:
Solve:
Solving a Polynomial Equation
x4  6x2  8x + 24  0.
Solution The graph of f(x)  x4  6x2  8x + 24 is shown the figure below.
Because the x-intercept is 2, we will test 2 by synthetic division and show that
it is a root of the given equation.
2
1
1
0 6 8 24
2
4 4 
24
2 2 12 The
0 zero remainder
indicates that 2 is a root
of x4  6x2  8x + 24  0.
x-intercept: 2
EXAMPLE:
Solve:
Solving a Polynomial Equation
x4  6x2  8x + 24  0.
Solution
Now we can rewrite the given equation in factored form.
x4  6x2  8x + 24  0
(x – 2)(x3  2x2  2x  12)  0
x–20
or
x3  2x2  2x  12  0
This is the given equation.
This is the result obtained from the
synthetic division.
Set each factor equal to zero.
Now we must continue by factoring x3 + 2x2 - 2x - 12 = 0
EXAMPLE:
Solve:
Solving a Polynomial Equation
x4  6x2  8x + 24  0.
Solution Because the graph turns around at 2, this means that 2 is a root of
even multiplicity. Thus, 2 must also be a root of x3  2x2  2x  12 = 0.
These are the coefficients
of x3  2x2  2x  12 = 0.
2
1
1
x-intercept: 2
2
2
4
2 12
8 12
6
0
The zero remainder
indicates that 2 is a root of
x3  2x2  2x  12 = 0.
EXAMPLE:
Solve:
Solution
Solving a Polynomial Equation
x4  6x2  8x + 24  0.
Now we can solve the original equation as follows.
x4  6x2  8x + 24  0
(x – 2)(x3  2x2  2x  12)  0
(x – 2)(x – 2)(x2  4x  6)  0
x – 2  0 or x – 2  0 or x2  4x  6  0
x2
x2
x2  4x  6  0
This is the given equation.
This was obtained from the first
synthetic division.
This was obtained from the second
synthetic division.
Set each factor equal to zero.
Solve.
EXAMPLE:
Solve:
Solution
Solving a Polynomial Equation
x4  6x2  8x + 24  0.
We can use the quadratic formula to solve x2  4x  6  0.
b  b2  4ac
x
2a
4  42  4 1  6 

2 1
4   8

2
4  2i 2

2
 2  i 2
We use the quadratic formula because x2  4x  6  0
cannot be factored.
Let a  1, b  4, and c  6.
Multiply and subtract under the radical.
8 
4(2)(1)  2i 2
Simplify.
The solution set of the original equation is {2, 2  i 2, 2  i 2}.
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