Finding the zeros of Polynomial Function - Math 4 Topics

1. Finding the Zeros of a
Polynomial Function
Reynaldo B. Pantino, T2

2. Objectives
1.) To determine the zeros of polynomial
functions of degree greater than 2 by;
a.) factor theorem
b.) factoring
c.) synthetic division
d.) depressed equations
2.)To determine the zeros of polynomial
functions of degree n greater than 2 expressed
as a product of linear factors.

3. Recapitulations
What is remainder theorem?
What is synthetic division?
What is factoring?
What is zero of a function?

4. Discussions
UNLOCKING OF DIFFICULTIES
The zero of a polynomial function P(x) is the
value of the variable x, which makes
polynomial function equal to zero or P(x) =
0.

5. Discussions
UNLOCKING OF DIFFICULTIES
The fundamental Theorem of Algebra states
that “Every rational polynomial function
P(x) = 0 of degree n has exactly n zeros”.

6. Discussions
UNLOCKING OF DIFFICULTIES
When a polynomial is expressed as a product
of linear factors, it is easy to find the zeros
of the related function considering the
principle of zero products.

7. Discussions
UNLOCKING OF DIFFICULTIES
The principle of zero product state that, for all
real numbers a and b, ab = 0 if and only if
a = 0 or b = 0, or both.

8. Discussions
UNLOCKING OF DIFFICULTIES
The degree of a polynomial function
corresponds to the number of zeros of the
polynomial.

9. Discussions
UNLOCKING OF DIFFICULTIES
A depressed equation of P is an equation
which has a degree less that of P.

10. Discussions
Illustrative Example 1
Find the zeros of
P(x) = (x – 3)(x + 2)(x – 1)(x + 1).
Solution: (Use the principle of zero products)
P(x) = 0; that is
x - 3 = 0 x + 2 = 0 x - 1 = 0 x + 1 = 0
x = 3 x = -2 x = 1 x = -1

11. Discussions
Illustrative Example 2
Find the zeros of
P(x) = (x + 1)(x + 1)(x +1)(x – 2)
Solution: (By zero product principle)
we have, P(x) = 0 the zeros are -1 and 2.
The factor (x + 1) occurs 3 times. In this case, the
zero -1 has a multiplicity of 3.

12. Discussions
Illustrative Example 3
Find the zeros of P(x) = (x + 2)3(x2 – 9).
Solution: (By factoring)
we have, P(x) = (x +2)(x+2)(x+2)(x – 3)(x + 3).
The zeros are;
-2, 3, -3,
where -2 has a multiplicity of 3.

13. Discussions
Illustrative Example 4
Function Zeros No. of
Zeros
P(x) = x – 4
P(x) = x2 + 8x + 15
P(x) = x3 -2x2 – 4x + 8
P(x) = x4 – 2x2 + 1
4 1
-3, -5 2
2, -2, 2 3
1,1,-1,-1 4

14. Discussions
Illustrative Example 4
Solve for the zeros of
P(x) = x3 + 8x2 + 19x + 12, given that one zero is -1.
Solution: By factor theorem, x + 1 is a factor of
x3 + 8x2 + 19x + 12.
Then; P(x) = x3 + 8x2 + 19x + 12
= (x+1)● Q(x).

15. Discussions
Illustrative Example 4 (Continuation of solution)
To determine Q(x), divide x3 + 8x2 + 19x + 12 by
(x + 1). By synthetic division;
--11 11 88 1199 1122
11
--11
77
--77
1122
--1122
00

16. Discussions
Illustrative Example 4 (Continuation of solution)
The equation x2 + 7x + 12 is a depressed
equation of P(x). To find the remaining zeros
use this depressed equation.
By factoring we have;
x2 + 7x + 12 = 0
(x +3)(x + 4) = 0
x = -3 and x = -4
Observe that a polynomial
function of degree 3 has
exactly three zeros.
Therefore; the three zeros are -1, -3, and -4.

17. Exercises
1. Solve for the other zeros of
P(x) = x4 – x3 – 11x2 + 9x + 18, given that one zero is -3.
2. Solve for the other zeros of
P(x) = x3 – 2x2 – 3x + 10, given that – 2 is a zero.

18. Activity Numbers
Which of the numbers -3, -2, -1, 0, 1, 2, 3 are
zeros of the following polynomials?
1.) f(x) = x3 + x2 + x + 1
2.) g(x) = x3 – 4x2 + x + 6
3.) h(x) = x3 – 7x + 6
4.) f(x) = 3x3 + 8x2 – 2x + 3
5.) g(x) = x3 + 3x2 – x – 3

19. Activity Factors
Which of the binomials (x – 1), (x + 1), (x – 4),
(x + 3) are factors of the given polynomials.
1.) x3 + x2 - 7x + 5
2.) 2x3 + 5x2 + 4x + 1
3.) 3x3 – 12x2 + 2x – 8
4.) 4x4 - x3 + 2x2 + x – 3
5.) 4x4 + 5x3 - 14x2 – 4x + 3

20. Activity Zeros
Find the remaining zeros of the polynomial
function, real or imaginary, given one of its
zeros.
1.) P(x) = x3 + 5x2 - 2x – 24 x = 2
2.) P(x) = x3 - x2 - 7x + 3 x = 3
3.) P(x) = x3 – 8x2 + 20x – 16x = 2
4.) P(x) = x3 + 5x2 - 9x – 45 x = -5
5.) P(x) = x3 + 3x2 + 3x + 1 x = -1

21. Assignments
On page 103, answers numbers 6, 12, 18,19, & 20.
Ref. Advanced Algebra, Trigonometry & Statistics
What is rational Zero Theorm? Pp. 105