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Reynaldo B. Pantino, T2

Finding the Zeros of a Polynomial Function

Objectives

1.) To determine the zeros of polynomial functions of degree greater than 2 by;

a.) factor theoremb.) factoringc.) synthetic divisiond.) depressed equations

2.)To determine the zeros of polynomial functions of degree n greater than 2 expressed as a product of linear factors.

Recapitulations

What is remainder theorem?

What is synthetic division?

What is factoring?

What is zero of a function?

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.

Discussions


The fundamental Theorem of Algebra states that “Every rational polynomial function

P(x) = 0 of degree n has exactly n zeros”.

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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.

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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.

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The degree of a polynomial function corresponds to the number of zeros of the polynomial.

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A depressed equation of P is an equation which has a degree less that of P.

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Illustrative Example 1Find 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 = 0x + 1 = 0

x = 3 x = -2 x = 1 x = -1

Discussions

Illustrative Example 2Find 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.

Discussions

Illustrative Example 3Find 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.

Discussions

Illustrative Example 4 Function Zeros No.

of Zero

sP(x) = x – 4

P(x) = x2 + 8x + 15P(x) = x3 -2x2 – 4x + 8

P(x) = x4 – 2x2 + 1

4 1

32, -2, 2

-3, -5 2

1,1,-1,-1 4

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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).

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Illustrative Example 4 (Continuation of solution)

To determine Q(x), divide x3 + 8x2 + 19x + 12 by

(x + 1). By synthetic division;-1-1 11 88 1919 1212

11-1-177

-7-71212

-12-1200

Discussions

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

Therefore; the three zeros are -1, -3, and -4.

Observe that a polynomial function of degree 3 has

exactly three zeros.

Illustrative Example 4 (Continuation of solution)

Exercises

1. Solve for the other zeros ofP(x) = x4 – x3 – 11x2 + 9x + 18, given that one zero is -3.

2. Solve for the other zeros ofP(x) = x3 – 2x2 – 3x + 10, given that – 2 is a zero.

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 + 12.) g(x) = x3 – 4x2 + x + 63.) h(x) = x3 – 7x + 6 4.) f(x) = 3x3 + 8x2 – 2x + 3 5.) g(x) = x3 + 3x2 – x – 3

Activity Factors

Which of the binomials (x – 1), (x + 1), (x – 4), (x + 3) are factors of the given polynomials.

1.) x3 + x2 - 7x + 52.) 2x3 + 5x2 + 4x + 13.) 3x3 – 12x2 + 2x – 8 4.) 4x4 - x3 + 2x2 + x – 3 5.) 4x4 + 5x3 - 14x2 – 4x + 3

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 = 22.) P(x) = x3 - x2 - 7x + 3 x = 33.) P(x) = x3 – 8x2 + 20x – 16 x = 24.) P(x) = x3 + 5x2 - 9x – 45 x = -55.) P(x) = x3 + 3x2 + 3x + 1 x = -1

Assignments

On page 103, answers numbers 6, 12, 18,19, & 20.Ref. Advanced Algebra, Trigonometry & Statistics

What is rational Zero Theorm? Pp. 105