Download The Fundamental Theorem of Algebra

Section 2.5
Objectives:
Use the Fundamental Theorem of
Algebra to determine the number of
zeros of a polynomial function.
 Find all zeros of polynomial functions,
including complex zeros.
 Find conjugate pairs of complex zeros.
 Find zeros of polynomials by factoring.

Fundamental Theorem of Algebra
The fundamental theorem of algebra
guarantees that every polynomial has a
complete factorization, if we are allowed to
use complex numbers.
The Linear Factorization Theorem

Using the Fundamental Theorem of
Algebra and the equivalence of zeros
and factors, we obtain the following
theorem.
Example 1
All zeros for the given polynomial are
distinct. Use the figure to determine
graphically the
number of real
zeros and the
number of
imaginary zeros.
f(x) = 3x3 – 3x2 – 3x – 5
Example 1
Solution
Real zeros correspond to x-intercepts;
imaginary zeros do not.
Graph has one xintercept so there is
one real zero.
Since f is degree 3
and all zeros are
distinct, there are two
imaginary numbers.
Polynomial Equations with
Complex Solutions
We can solve polynomial equations by
writing them in their complete factored
form and then use the fact that the
solutions to the equation are the zeros of
f(x).
Example 2
Confirm that the third-degree polynomial function
f(x) = x3 – x2 + 4x – 4 has exactly three zeros: x =
1, x = 2i, and x = -2i.
Solution
 Factor f(x) by grouping

x3 – x2 + 4x – 4
(x3 – x2) + (4x – 4)
x2(x-1) + 4(x-1)
(x-1)(x2+4)

Solve for the zeros (zero product property)
x-1 = 0
x=1
x2+4 = 0
x2 = -4
x = ±√4
x = 2i, -2i
Your Turn:

Confirm that the third-degree polynomial
function f(x) = x3 + 4x2 + 9x + 36 has exactly
three zeros: x = -4, x = 3i, and x = -3i.
Solution
 Factors (x+4)(x-3i)(x+3i)
 Zeros x = -4, 3i, -3i
Example 3a:
Find all zeros of f(x) = x3 + 3x2 + 16x + 48





(should be 3 total!)
p = ±1 ±2 ±3 ±4 ±6 ±8 ±12 ±16 ±24 ±48
q
±1
Graph and you’ll see -3 is the only one on the graph you can see so
synthetic divide with -3
-3
1

1
3
-3
16
0
0
16
48
-48
0
x2 + 16 = 0
x2 = -16
x = ±√-16 = ±4i
The three zeros are -3, 4i, -4i
Example 3b

Write f(x) = x3 + 3x2 + 16x + 48 as a
product of linear factors.
 Zeros are 3, 4i, -4i
 Zero 3 ⇒ factor (x-3)
 Zero 4i ⇒ factor (x-4i)
 Zero -4i ⇒ factor (x+4i)

f(x) = (x-3)(x-4i)(x+4i)
Example 4
Solve x3 - 3x2  7x + 21.
Solution
Rewrite the equation as f(x) = 0, where
f(x) = x3  3x2 + 7x  21.
We can use factoring by grouping to find one
real zero of f(x).
Example 3
Solution continued
x3  3x2 + 7x  21
(x3 – 3x2) + (7x – 21)
x2(x – 3) + 7(x – 3)
(x – 3)(x2 + 7)
x3  3x2 + 7x  21 = (x  3)(x2 + 7)
x 3 0
or
x 3
or
x 3
or
x2  7  0
x  7
2
x  i 7
The solutions are 3 and x  i 7.
Your Turn:

Write f(x) = x5 -3x4 -7x3 + 21x2 -18x -54 as
the product of linear factors, and list all the
zeros of f.
Solution
 Factors: f(x) = (x+3)(x-3)(x-3)(x-i√2)(x+i√2)
 Zeros: x = 3, -3, -3, i√2, -i√2
Conjugate Zeros Theorem
If a polynomial f(x) has only real
coefficients and if a + bi is a zero of
f(x), then the conjugate a – bi is
also a zero of f(x).
Complex Zeros Occur in Conjugate Pairs
Zeros of a Polynomial
Zeros (Solutions)
Real Zeros
Complex Zeros
Rational Zeros
Complex Number
and its Conjugate
Summary of Theorems
Fundamental Theorem of Algebra
Every polynomial P(x) of degree n > 0 has at least one zero.
Linear Factorization Theorem
Every polynomial P(x) of degree n > 0 can be expressed as the product of
n linear factors. Hence, P(x) has exactly n zeros—not necessarily distinct.
Complex Conjugates Zeros Theorem
Imaginary zeros of polynomials with real coefficients, if they exist, occur
in conjugate pairs.
Real Zeros and Odd-Degree Polynomials
A polynomial of odd degree with real coefficients always has at least one
real zero.
Solution
f(x) = (x – 3)(x + 1)(x – 2i)(x + 2i)
Multiply these to get the polynomial
f(x) = (x2 – 2x – 3)(x2 + 4)
f(x) = x4 – 2x3 + x2 – 8x – 12
Example 5
Represent a polynomial f(x) of degree 4 with
leading coefficient 2 and zeros of 3, 5, i and
i in complete factored form and expanded
form.
Solution
Let an = 2, x1 = 3, x2 = 5, x3 = i, and x4 = i.
f(x) = 2(x + 3)(x  5)(x  i)(x + i)
Expanded: 2(x + 3)(x  5)(x  i)(x + i)
= 2(x + 3)(x  5)(x2 + 1)
= 2(x + 3)(x3  5x2 + x  5)
= 2(x4  2x3  14x2  2x  15)
= 2x4  4x3 – 28x2 – 4x – 30
Your Turn:

Find a fourth – degree polynomial
function with real coefficients that has 2, -2, and 2i as zeros.
Solution
f(x) = x4 + 4x3 + 8x2 + 16x + 16
Example 6
Find the zeros of f(x) = x4 + x3 + 2x2 + x + 1
given one zero is i.
Solution
By the conjugate zeros theorem it follows that i
must be a zero of f(x).
(x + i) and (x  i) are factors
(x + i)(x  i) = x2 + 1, using long division we
can find another quadratic factor of f(x).
Example 6
Solution continued
Long division
x2  x  1
x 2  0 x  1 x 4  x 3  2x 2  x  1
x 4  0x3  x 2
x3  x2  x
x3  0x 2  x
x 2  0x  1
x 2  0x  1
0
So
x4 + x3 + 2x2 + x + 1 = (x2 + 1)(x2 + x + 1)
Example 6
Solution continued
Use the quadratic formula to find the zeros
of x2 + x + 1.
b  b 2  4ac
x
2a
x
1  12  4 11
2 1
1
3
x  i
2
2
The four zeros are:
1
3
i and   i
2
2
Example 7: Now write a polynomial function of least
degree that has real coefficients, a leading
1 and 1, -2+i, -2-i as zeros.









F(x)= (x-1)(x-(-2+i))(x-(-2-i))
F(x)= (x-1)(x+2-i)(x+2+i)
f(x)= (x-1){(x+2)-i} {(x+2)+i}
F(x)= (x-1){(x+2)2-i2}
Foil
F(x)=(x-1)(x2 + 4x + 4 –(-1))
Take care
of i2
F(x)= (x-1)(x2 + 4x + 4 + 1)
F(x)= (x-1)(x2 + 4x + 5)
Multiply
F(x)= x3 + 4x2 + 5x – x2 – 4x – 5
f(x)= x3 + 3x2 + x - 5
coeff. of
Example 8: Now write a polynomial function of least degree
that has real coefficients, a leading coeff. of
1 and 4, 4, 2+i as zeros.









Note: 2+i means 2-i is also a zero
F(x)= (x-4)(x-4)(x-(2+i))(x-(2-i))
F(x)= (x-4)(x-4)(x-2-i)(x-2+i)
F(x)= (x2 – 8x +16)((x-2)-i)((x-2)+i)
F(x)= (x2 – 8x +16)((x-2)2-i2)
F(x)= (x2 – 8x +16)(x2 – 4x + 4 –(-1))
F(x)= (x2 – 8x +16)(x2 - 4x + 5)
F(x)= x4–4x3+5x2–8x3+32x2-40x+16x2-64x+80
F(x)= x4-12x3+53x2-104x+80
Example 9
4
3
2
Find all complex zeros of P( x)  x  7 x  18 x  22 x  12
given that 1 – i is a zero.
Solution
1 i 1
1
7
18
 22
12
1  i  7  5i
16  6i  12
 6  i 11  5i
 6  6i
0
Example 9
Using the Conjugate Zeros Theorem, 1 + i is also a
zero.
1 i 1
 6  i 11  5i
 6  6i
1  i  5  5i
6  6i
1
5
6
0
The zeros of x2 – 5x + 6 are 2 and 3. Thus,
P( x)  x 4  7 x3  18 x 2  22 x  12
 ( x  2)( x  3)( x  1  i )( x  1  i )
and has four zeros: 1 – i, 1 + i, 2, and 3.
Your Turn:
Write as a product of linear factors:
f(x) = x4 – 16
f(x) = (x-2)(x+2)(x-2i)(x+2i)
 Find a third-degree polynomial with integer
coefficients that has 2 and 3-I as zeros.
f(x) = x3 – 8x2 + 22x – 20
 Write the polynomial f(x) = x4 – 4x3 + 5x2 –
2x – 6 in completely factored form. (Hint:
one factor is x2 – 2x – 2)
