Zeros of Polynomial Functions Lesson 2-4

Objectives:

1.Use the Fundamental Theorem of Algebra to determine the number of zeros of polynomial functions2.Find rational zeros of polynomial functions3.Find conjugate pairs of complex zeros4.Find zeros of polynomials by factoring5.Use Descartes’s Rule of Signs and the Upper and Lower Bound Rules to find zeros of polynomials

In the complex number system, every nth-degree polynomial has precisely “n” zeros.

Fundamental Theorem of Algebra

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

Linear Factorization Theorem

If f(x) is a polynomial of degree n, where n > 0, then f has precisely n linear factors

where are complex numbers

f (x) = an (x − c1)(x − c2)...(x − cn )

c1,c2,....,cn

Zeros of Polynomial Functions

Give the degree of the polynomial, tell how many zeros there are, and find all the zeros

f (x) = x − 2

f (x) = x 2 − 6x + 9

f (x) = x 3 + 4x

f (x) = x 4 −1

Rational Zero TestTo use the Rational Zero Test, you should list all rational numbers whose numerators are factors of the constant term and whose denominators are factors of the leading coefficient

Once you have all the possible zeros test them using substitution or synthetic division to see if they work and indeed are a zero of the function (Also, use a graph to help determine zeros to test)

It only test for rational numbers

Possible_Rational_Zeros=factors_of _constant _ term

factors_of _ leading_coefficient

EXAMPLE: Using the Rational Zero Theorem

List all possible rational zeros of f (x) =

15x3 +

14x2 −

3x – 2.

Solution The constant term is –2 and the leading coefficient is 15.

1 2 1 2 1 25 53 3 15 15

Factors of the constant term, 2Possible rational zeros Factors of the leading coefficient, 151, 2

1, 3, 5, 151, 2, , , , , ,

−=

± ±=± ± ± ±

= ± ± ± ± ± ± ± ±

Divide ±1 and ±2 by ±1.

Divide ±1 and ±2 by ±3.

Divide ±1 and ±2 by ±5.

Divide ±1 and ±2 by ±15.

There are 16 possible rational zeros. The actual solution set to f (x) =

15x3 +

14x2 −

3x – 2 = 0 is {-1, −1/3, 2/5}, which contains 3 of the 16 possible solutions.

Roots & Zeros of Polynomials II

Finding the Roots/Zeros of Polynomials (Degree 3 or higher):

• Graph the polynomial to find your first zero/root

• Use synthetic division to find a smaller

polynomial

• If the polynomial is not a quadratic follow the 2

steps above using the smaller polynomial until

you get a quadratic.

• Factor or use the quadratic formula to find your

Example 1:

Find all the zeros of each polynomial function

3 210 9 19 6x x x+ − + First, graph the equation to find the first zero

From looking at the graph you can see that there is a zero at -2

ZERO

Example 1 Continued3 210 9 19 6x x x+ − + Second, use the zero you

found from the graph and do synthetic division to find a smaller polynomial

-2 10 9 -19 6 Don’t forget your remainder should be zero-20 22 -6

10 -11 3 0

The new, smaller polynomial is:

210 11 3x x− +

Example 1 Continued:

210 11 3x x− + Third, factor or use the quadratic formula to find the remaining zeros.

This quadratic can be factored into: (5x – 3)(2x – 1)

Therefore, the zeros to the problem are:3 210 9 19 6x x x+ − +

3 12, ,5 2

x = −

Rational ZerosFind the rational zeros.

f (x) = x 3 + x +1

f (x) = x 4 − x 3 + x 2 − 3x − 6

f (x) = x 3 + 8x 2 + 40x − 525

f (x) = 2x 3 + 3x 2 − 8x + 3

Find all the real zeros (Hint: start by finding the rational zeros)

f (x) = −10x 3 +15x 2 +16x −12

f (x) = 3x 3 −19x 2 + 33x − 9

Writing a Polynomial given the zeros.

To write a polynomial you must write the zeros out in factored form. Then you multiply the factors together to get your polynomial.

Factored Form: (x – zero)(x – zero). . .

***If P is a polynomial function and a + bi is a root, then a – bi is also a root.***If P is a polynomial function and is a root, then is also a root

a b+a b−

Example 1:

The zeros of a third-degree polynomial are 2 (multiplicity 2) and -5. Write a polynomial.

(x – 2)(x – 2)(x – (-5)) = (x – 2)(x – 2)(x+5)

First, write the zeros in factored form

Second, multiply the factors out to find your polynomial

Example 1 Continued(x – 2)(x – 2)(x+5)

2( 2)( 2) 4 4x x x x− − = − + First FOIL or box two of the factors

2 4 4x x− +X

5

3x 24x− 4x25x 20x− 20

Second, box your answer from above with your remaining factors to get your polynomial:

3 2 16 20x x x+ − +ANSWER

So if asked to find a polynomial that has zeros, 2 and 1 – 3i, you would know another root would be 1 + 3i. Let’s find such a polynomial by putting the roots in factor form and multiplying them together.

( ) ( ) ( )ixixx 31312 −−+−− Multiply the last two factors together. All i terms should disappear when simplified.

( ) ( )22 9333132 iixiixxixxx −−+++−−−−-1( ) ( )1022 2 +−−= xxx Now multiply the x – 2 through

20144 23 −+−= xxxHere is a 3rd degree polynomial with roots 2, 1 - 3i and 1 + 3i

If x = the root then x - the root is the factor form. ( ) ( )( ) ( )( )ixixx 31312 +−−−−

Conjugate PairsComplex Zeros Occur in Conjugate Pairs = If a + bi is a zero of the function, the conjugate a – bi is also a zero of the function (the polynomial function must have real coefficients)

EXAMPLES: Find a polynomial with the given zeros

-1, -1, 3i, -3i

2, 4 + i, 4 – i

STEPS For Finding the Zeros given a Solution

1) Find a polynomial with the given solutions (FOIL or BOX)

2) Use long division to divide your polynomial you found in step 1 with your polynomial from the problem

3) Factor or use the quadratic formula on the answer you found from long division.

4) Write all of your answers out

Find Roots/Zeros of a PolynomialIf the known root is imaginary, we can use the Complex Conjugates Thm.

Ex: Find all the roots of f (x) = x3 − 5x2 − 7x + 51

If one root is 4 - i.

Because of the Complex Conjugate Thm., we know that another root must be 4 + i.

Example (con’t)Ex: Find all the roots of f (x) = x3 − 5x2 − 7x + 51

If one root is 4 - i.

If one root is 4 - i, then one factor is [x - (4 - i)], and

Another root is 4 + i, & another factor is [x - (4 + i)].

Multiply these factors:[ (4 )][ (4 )] ( 4 )( 4 )x i x i x i x i− − − + = − + − −

4x i− +X

-4

-i

2x 4x− ix

4x− 16 4i−

ix− 4i 2i−

2 8 17x x= − +

Example (con’t)Ex: Find all the roots of f (x) = x3 − 5x2 − 7x + 51

If one root is 4 - i.

x2 − 8x + 17

If the product of the two non-real factors is

x2 − 8x + 17then the third factor (that gives us the real root) is the quotient of P(x) divided by

x2 − 8x +17 x3 − 5x2 − 7x + 51x3 − 5x2 − 7x + 51 0

x +3) The third

root is x = -3

So, all of the zeros are: 4 – i, 4 + i, and -3

FIND ALL THE ZEROS

f (x) = x 4 − 3x 3 + 6x 2 + 2x − 60

(Given that 1 + 3i is a zero of f)

f (x) = x 3 − 7x 2 − x + 87

(Given that 5 + 2i is a zero of f)

More Finding of Zerosf (x) = x 5 + x 3 + 2x 2 −12x + 8

f (x) = 3x 3 − 4x 2 + 8x + 8

Descartes’s Rule of Signs

Let ; be a polynomial with real coefficients and

The number of positive real zeros of f is either equal to the number of variations in sign of f(x) or less than that number by an even integerThe number of negative real zeros of f is either equal to the number of variations in sign of f(-x) or less than that number by an even integer

Variation in sign = two consecutive coefficients have opposite signs

f (x) = an xn + an−1xn−1 + ...+ a2x

2 + a1x + a0 a0 ≠ 0

EXAMPLE: Using Descartes’ Rule of Signs

Determine the possible number of positive and negative real zeros of f (x) =

x3 +

2x2 + 5x + 4.

Solution1. To find possibilities for positive real zeros, count the number of sign

changes in the equation for f (x). Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros.

2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f (−x). We obtain this equation by replacing x with −x in the given function.

f (−x) =

(−x)3 +

2(−x)2 + 5(−x) + 4

f (x) =

x3 +

2x2 + 5x + 4 This is the given polynomial function.

Replace x with −x.

= −x3 +

2x2 −

5x + 4

EXAMPLE: Using Descartes’ Rule of Signs

Determine the possible number of positive and negative real zeros of f (x) =

x3 +

2x2 + 5x + 4.

SolutionNow count the sign changes.

There are three variations in sign. # of negative real zeros of f is either equal to 3, or is less than this number by an even integer. This means that there are either 3 negative real zeros

or 3 −

2 =

1 negative real zero.

f (−x) = −x3 +

2x2 −

5x + 4

1 2 3

Descartes’s Rule of Signs EXAMPLES: describe the possible real zeros

f (x) = 3x 3 − 5x 2 + 6x − 4f (x) = 3x 3 + 2x 2 + x + 3

Upper & Lower Bound Rules

Let f(x) be a polynomial with real coefficients and a positive leading coefficient. Suppose f(x) is divided by x – c, using synthetic didvision

If c > 0 and each number in the last row is either positive or zero, c is an upper bound for the real zeros of fIf c < 0 and the numbers in the last row are alternately positive and negative (zero entries count as positive or negative), c is a lower bound for the real zeros of f

EXAMPLE: find the real zeros f (x) = 6x 3 − 4x 2 + 3x − 2

h(x) = x4 + 6x3 + 10x2 + 6x + 9

1 1 6 10 6 9

1

2 4

6

6

0

0

4 9Signs are all positive, therefore 1 is an

upper bound.

1 of Factors9 of Factors

1931

±±±±

=,,

EXAMPLEYou are designing candle-making kits. Each kit contains 25 cubic inches of candle wax and a model for making a pyramid- shaped candle. You want the height of the candle to be 2 inches less than the length of each side of the candle’s square base. What should the dimensions of your candle mold be?