This document provides examples of using the Fundamental Theorem of Algebra to find all zeros of a polynomial function given some of the zeros. It demonstrates finding the remaining complex zeros by using the conjugate pairs theorem. The last example solves a polynomial function of degree 4 using the quadratic formula to find the two complex zeros.

1. Section 5.6
Complex Zeros;
Fundamental Theorem of
Algebra

6. Example 1
A polynomial of degree 5 whose coefficients are real numbers
has the zeros -2, -3i, and 2+4i. Find the remaining two zeros.

7. By the conjugate pairs theorem, if x = -3i and x = 2+4i
then x = 3i and x = 2-4i

8. Example 2: Find the remaining zeros of f.
Degree 5; zeros: 1, i, 2i
The degree
We have three
indicates the
zeros, so we are
number of zeros a
missing two!!!
polynomial has.
By the Conjugate
Pairs Theorem Remaining zeros: -i, -2i

10. Example 3
Find a polynomial f of degree 4 whose coefficients are
real numbers and that has the zeros 1, 1, 4+i.

11. By the Conjugate
Pairs Theorem
x = 1, x = 1, x = 4 + i, x = 4 − i
P ( x) = ( x − 1)( x − 1)( x − 4 − i )( x − 4 + i )
P( x) = ( x − x − x + 1)( x − 4 x + xi − 4 x +
2 2
16 − 4i − xi + 4i − i ) 2
P( x) = ( x − 2 x + 1)( x − 8 x + 16 − (−1))
2 2
P( x) = ( x − 2 x + 1)( x − 8 x + 17)
2 2

12. P( x) = x − 8 x + 17 x − 2 x + 16 x −
4 3 2 3 2
34 x + x − 8 x + 17
2
P( x) = x − 10 x + 34 x − 42 x + 17
4 3 2

15. Example 4: Use the given zero to find the remaining
zeros of each function.
f ( x) = x − 4 x + 4 x − 16; zero : 2i
3 2
2i 1 -4 4 - 16
2i - 4 − 8i 16
1 - 4 + 2i - 8i 0
Note:
2i (-4 + 2i ) = -8i + 4i = -8i + 4(-1) = -4 − 8i
2
2i (-8i ) = -16i = -16(-1) = 16
2

16. - 2i 1 - 4 + 2i - 8i
- 2i 8i
1 -4 0
4 1 -4
4
1 0
Zeros: 2i, -2i, 4

17. Example 5
Find the complex zeros of the polynomial function
x 4 + 2 x3 + x 2 − 8 x − 20 ±1, ± 2, ± 4, ± 5, ± 10, ± 20

18. 2 1 2 1 -8 - 20
2 8 18 20
1 4 9 10 0
-2 1 4 9 10
-2 -4 - 10
1 2 5 0
x + 2x + 5 = 0
2 +5
Note: The resulting quadratic equation can not be
factored since there is no number that multiplied gives
you five and at the same time added gives you two.

19. We have to use completing the square or the quadratic
formula 2
x + 2x + 5 = 0
a =1 b=2 c=5
Quadratic Formula
− b ± b − 4ac
2
x=
2a
Substitute − (2) ± (2) − 4(1)(5)
2
x=
2(1)

20. Simplify and solve!
− 2 ± 4 − 20
x=
2
− 2 ± − 16
x=
2
− 2 ± 4i
x= = −1 ± 2i
2
Complex zeros: x = −1± 2i
Real zeros: x = ±2
Remember the problem only asks for the complex zeros!