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2.5 Zeros of Polynomial Functions. Fundamental Theorem of Algebra Rational Zero Test Upper and Lower bound Rule. Fundamental Theorem of Algebra. If f(x) is a polynomial of degree “n” > 0, then f(x) has at least one zero in the complex number system. Complex zero’s (roots) come in pairs

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2 5 zeros of polynomial functions

2.5 Zeros of Polynomial Functions

Fundamental Theorem of Algebra

Rational Zero Test

Upper and Lower bound Rule


Fundamental theorem of algebra
Fundamental Theorem of Algebra

If f(x) is a polynomial of degree “n” > 0, then f(x) has at least one zero in the complex number system.

Complex zero’s (roots) come in pairs

If a + bi is a zero, then a – bi is a zero.


Linear factorization theorem
Linear Factorization Theorem

If f(x) is a polynomial of degree “n”>0, then there are as many zeros as degree.

If f(x) is a third degree function, then

f(x) = an(x – c1)(x – c2)(x – c3) where care complex numbers.

Complex zero’s (roots) come in pairs

If a + bi is a zero, then a – bi is a zero.


The rational zero test
The Rational Zero Test

If f(x) has integer coefficients, then all possible zeros are

factors of the constant

factor of the lead coefficient


The rational zero test1
The Rational Zero Test

If f(x) has integer coefficients, then all possible zeros are

factors of the constant

factor of the lead coefficient

f(x) = x 3 – 7x 2 + 4x + 12

Possible zeros ± 1, ± 2, ± 3, ± 4, ± 6, ± 12

± 1


F x x 3 7x 2 4x 12 possible zeros 1 2 3 4 6 12 1
f(x) = x 3 – 7x 2 + 4x + 12Possible zeros ± 1, ± 2, ± 3, ± 4, ± 6, ± 12 ± 1

- 1 | 1 -7 4 12

-1 8 -12

1 - 8 12 0

So – 1 is a zero

How do you want to find the other zeros.

x 2 – 8x + 12


Find the zeros f x 3x 3 x 2 6x 2
Find the zeros f(x) = 3x3 – x2 + 6x - 2


Descartes rule of signs
Descartes' Rule of Signs

Let f(x) = anxn + an-1xn-1 + ….a1x + a0 ; with real coefficients and a0≠ 0.

Part 1

The number of positive real zeros equals (or a even number less), the number of variation in the sign of the coefficient (switching from positive to negative or negative to positive).


Descartes rule of signs1
Descartes' Rule of Signs

Let f(x) = anxn + an-1xn-1 + ….a1x + a0 ; with real coefficients and a0≠ 0.

Part 2

The number of negative real zeros equals (or a even number less), the number of variation in the sign of the coefficient

(switching from positive to negative or negative to positive) in f(- x).


Using the desecrate rule of signs
Using the Desecrate rule of signs

f(x) = 4x3 - 3x2 +2x – 1

How many times does the sign change ?


Using the desecrate rule of signs1
Using the Desecrate rule of signs

f(x) = 4x3 - 3x2 +2x – 1

How many times does the sign change ?

3 times.

There are 3 or 1 positive zeros.


Using the desecrate rule of signs2
Using the Desecrate rule of signs

f(x) = 4x3 - 3x2 +2x – 1

What about f( -x) = -4x3 – 3x2 – 2x - 1

How many times does the sign change ?


Using the desecrate rule of signs3
Using the Desecrate rule of signs

f(x) = 4x3 - 3x2 +2x – 1

What about f( -x) = -4x3 – 3x2 – 2x - 1

How many times does the sign change ?

No change, no negative zeros.


Upper and lower bound rule
Upper and Lower bound Rule

If c > 0 ( “c” the number you divide by) and the last row of synthetic division is all positive or zero, the c| is the upper bound

So there is no zero larger then c, where c > 0.

If c < 0 and the last row alternate signs

( zero count either way), then c is the lower bound.


F x 2x 3 5x 2 12x 5
f(x) = 2x3 – 5x2 + 12x - 5

Check to see if 3 is the upper bound?

3| 2 - 5 12 - 5 All signs are 6 3 45 positive.

2 1 15 40

3 is an upper bound


F x 2x 3 5x 2 12x 51
f(x) = 2x3 – 5x2 + 12x - 5

Check to see if - 1 is the lower bound?

- 1| 2 - 5 12 - 5 All signs are -2 7 -19 switch.

2 - 7 19 -24

-1 is an lower bound


F x 2x 3 5x 2 12x 52
f(x) = 2x3 – 5x2 + 12x - 5

Find the zeros


Homework
Homework

Page 160 – 164

# 5, 15, 23, 35,

42, 50, 57, 65,

73, 81, 85, 93,

103, 108, 111


Homework1
Homework

Page 160 – 164

# 9, 19, 29, 41,

53, 61, 64, 77,

87, 97, 105,125


One more time
One more time

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