P.o.D. – Use the Quadratic Formula to
solve each equation. Exact answers only
in lowest, reduced form.
1.) 𝑥 2 − 2𝑥 + 2 = 0
2.) 4𝑥 2 + 16𝑥 + 17 = 0
3
3.) 𝑥 2 − 6𝑥 + 9 = 0
2
1.) 1 ± 𝑖
1
2.) −2 ± 𝑖
2
3.) 2 ± 𝑖 √2
2.5 – Zeros of Polynomial Functions
The Fundamental Theorem of Algebra:
If f(x) is a polynomial of degree n, where
n>0, then f has at least one zero in the
complex number system.
Similarly stated, there will be the same
number of complex solutions as the
highest exponent, or degree, of the
polynomial.
Linear Factorization Theorem:
If f(x) is a polynomial of degree n where
n>0, then f has precisely n linear factors
𝑓(𝑥 ) = 𝑎𝑛 (𝑥 − 𝑐1 )(𝑥 − 𝑐2 ) … (𝑥 − 𝑐𝑛 )
where 𝑐1 , 𝑐2 , … 𝑐𝑛 are complex numbers.
*This is very similar to the Fundamental
Theorem of Algebra.
EX: Find all the zeros of each function.
a.) F(x)= x-4
b.) 𝑓(𝑥 ) = 𝑥 2 − 4𝑥 + 4
c.) 𝑓(𝑥 ) = 𝑥 3 + 9𝑥
d.) 𝑓(𝑥 ) = 𝑥 4 − 16
a.) Set the factor equal to zero and
solve.
x-4=0  x=4
b.) We first need to factor the
polynomial. We can do this by DeFOIL-ing
𝑓 (𝑥 ) = 𝑥 2 − 4𝑥 + 4 = (𝑥 − 2)(𝑥 − 2)
Now set each factor equal to zero.
𝑥−2=0→𝑥 =2
This is known as a double root, because
the same answer occurs twice.
c.) Again, we must factor.
𝑓(𝑥 ) = 𝑥 3 + 9𝑥 = 𝑥(𝑥 2 + 9)
Set each factor equal to zero.
𝑥=0
𝑥2 + 9 = 0 →
𝑥 2 = −9 →
𝑥 = √−9 →
𝑥 = ±3𝑖
X = 0, 3i, and -3i
Notice that we have a degree of 3 and 3
complex solutions. This agrees with the
Fundamental Theorem of Algebra.
d.) We must factor once again. This
time we will factor using the
Difference of Squares.
𝑓(𝑥 ) = 𝑥 4 − 16 = (𝑥 2 )2 − 42
= (𝑥 2 + 4)(𝑥 2 − 4)
= (𝑥 2 + 4)(𝑥 − 2)(𝑥 + 2)
Next, set each factor equal to zero. This
is known as the Zero Product Property.
𝑥 2 + 4 = 0 x-2=0
→ 𝑥 2 = −4 x=2
→ 𝑥 = √−4
→ 𝑥 = ±2𝑖
X+2=0
X= -2
X = 2i, -2i, 2, -2
EX: Find the rational zeros of
𝑓(𝑥 ) = 𝑥 3 − 5𝑥 2 + 2𝑥 + 8.
Recall, that the word “rational” means
fractional. Therefore, the only solutions
which we need to find are real,
fractional answers. This can easily be
done graphically.
x= -1,2,4
EX: Find the rational zeros of
𝑓(𝑥 ) = 𝑥 3 − 15𝑥 2 + 75𝑥 − 125
x=5
Since there is only one rational zero, we
know that there must be two imaginary
solutions (or a triple root) due to the
Fundamental Theorem of Algebra.
EX: Find the rational zeros of
𝑓(𝑥 ) = 2𝑥 4 − 9𝑥 3 − 18𝑥 2 + 71𝑥 − 30
The solutions are x= ½, -3, 2, 5.
EX: Find all the real solutions of
𝑓(𝑥 ) = 𝑥 3 − 7𝑥 2 − 11𝑥 + 14
x = -2, .85995, 8.14005
We can also solve polynomials
algebraically.
EX: Solve the previous problem
algebraically.
Step 1: Use any known method (typically
graphing) to find at least one real,
rational solution.
From the previous example, we know
that x= -2 is a solution. This means that
(x+2) is a factor.
Step 2: Use this factor to divide the
polynomial. We can use synthetic
division if we want.
-2
1
-7
-11
14
-2
18
-14
----------------------------------1
-9
7
0
This means that our other factor is 𝑥 2 −
9𝑥 + 7.
Step 3: Solve this factor set equal to zero
using the quadratic formula.
9 ± √(−9)2 − 4(1)(7) 9 ± √53
𝑥=
=
2(1)
2
Thus, our three solutions are
9±√53
2
and
-2.
Complex (Imaginary) Zeros Occur in
Conjugate Pairs:
Let f(x) be a polynomial function that has
real coefficients. If a+bi is a zero, then
the conjugate a-bi is also a zero of the
function.
This means that if 3-4i is a solution, then
3+4i must also be a solution.
EX: Find a third degree polynomial
function with integer coefficients that
has 2, 7i, and -7i as zeros.
Step 1: Write all possible binomial
factors.
(x-2)(x-7i)(x+7i)
Step 2: Expand (or multiply, FOIL,
distribute) the factors. We will do the
Difference of Squares’ factors first.
(𝑥 − 2)(𝑥 2 + 7𝑥𝑖 − 7𝑥𝑖 − 49𝑖 2 )
= (𝑥 − 2)(𝑥 2 − 49(−1))
= (𝑥 − 2)(𝑥 2 + 49)
= 𝑥 3 + 49𝑥 − 2𝑥 2 − 98
= 𝑥 3 − 2𝑥 2 + 49𝑥 − 98
EX: Find all the zeros of
𝑓(𝑥 ) = 𝑥 3 − 4𝑥 2 + 21𝑥 − 34, given
that 1+4i is a zero of f.
Step 1: Find a real, rational zero using
any known method.
x=2, so (x-2) is a factor.
Step 2: Divide f by the factor we just
found.
(show the synthetic division on the
whiteboard).
(𝑥 3 − 4𝑥 2 + 21𝑥 − 34) ÷ (𝑥 − 2)
= 𝑥 2 − 2𝑥 + 17
Step 4: Solve this factor set equal to
zero.
2 ± √(−2)2 − 4(1)(17)
𝑥=
2(1)
2 ± √−64 2 ± 8𝑖
=
=
= 1 ± 4𝑖
2
2
The three complex zeros are 2, 1+4i, and
1-4i.
EX: Write
ℎ(𝑥 ) = 𝑥 3 − 11𝑥 2 + 41𝑥 − 51 as the
product of linear factors, and list all of its
zeros.
Begin by solving the function.
(show all of the work on the whiteboard)
X = 3, 4+i, 4-i
Since these are the zeros, the factors
must be (x-3), (x-4-i), and (x-4+i).
Descartes’ Rule of Signs:
1. The number of positive real zeros of
a function is equal to the number of
sign changes of f(x) or by an even
number less.
2. The number of negative real zeros
of a function is equal to the number
of sign changes of f(-x) or by an
even number less.
EX: Describe the possible real zeros of
𝑓(𝑥 ) = −2𝑥 3 + 5𝑥 2 − 𝑥 + 8.
+1
+1
+1
There are three sign changes of f(x). This
means that there are 3 or 1 positive real
zero.
𝑓(−𝑥 ) = −2(−𝑥)3 + 5(−𝑥)2 − (−𝑥 ) +
8 = 2𝑥 3 + 5𝑥 2 + 𝑥+8
There are 0 sign changes in f(-x), so there
are 0 possible negative real zeros.
We can make a chart explaining this
situation.
Possible (+) Possible (-)
Imaginary
real zeros
real zeros
solutions
3
0
0
1
0
2
Let’s examine the graph to determine which
situation exists.
This graph verifies that we
only have 1 positive, real zero. Now find
that zero.
x = 2.8243985
http://www.youtube.com/watch?v=G7T1EY9i44Y
EX: Use Descartes’ Rule of Signs to
determine the possible real zeros of 𝑓(𝑥 ) =
8𝑥 3 − 4𝑥 2 + 6𝑥 − 3. Then find all of the
real, rational zeros.
There are 3 sign changes in f(x), so there are
3 or 1 possible positive, real zeros.
𝑓(−𝑥 ) = 8(−𝑥)3 − 4(−𝑥)2 + 6(−𝑥 ) − 3
= −8𝑥 3 − 4𝑥 2 − 6𝑥 − 3
There are 0 sign changes in f(-x), so there
are 0 possible negative, real zeros.
x=½
Upon completion of this lesson, you should
be able to:
1. Find all zeros of a polynomial function.
2. Apply Descartes’ Rule of Signs.
For more information, visit
http://www.purplemath.com/modules/drofsign.htm
HW Pg. 179 6-96 6ths, 106, 113, 129-132