Section 2-3 Real Zeros of Polynomial Functions. Real Zeros of Polynomial Functions What you should know: 1. How to divide polynomials by other polynomials

Section 2-3

Real Zeros of

Polynomial Functions

Real Zerosof Polynomial Functions

What you should know:

1. How to divide polynomials by other polynomials using long division.

2. How to divide polynomial by binominals in the form (x – k) usingsynthetic division and how to recognize upper and lower bounds.

3. How to apply the Remainder and Factor Theorems.

4. How to determine possible rational zeros of polynomial functions usingthe Rational Zero Test.

5. How to determine the number of positive and negative zeros a givenfunction has by applying Descartes Rule of Signs.

Real Zeros of Polynomial Functions

Long Division of Polynomials

To perform long division of polynomial functions, follow the same steps as onewould as if dividing a two-digit number into a three-digit number.

Remember the parts to a division problem.

quotientdivisor dividend

remainder

Example: Divide 3 26 19 16 4 by 2x x x x

3 22 6 19 16 4x x x x 3 26 12x x

27 16x x

2 4x 2 4x

0

27 14x x

6x2 – 7x + 2

Real Zeros of Polynomial Functions

2Divide 3 5 by 1x x x

21 3 5x x x 2x x

Try this:Label the parts.

2 5x 2 2x

3

x 2Divisor

Quotient

Dividend

Remainder

Write the result as f(x) = (divisor)(quotient) + remainder

f(x) = (x + 1)(x + 2) + 3


The Division Algorithm

If f(x) and d(x) are polynomials such that d(x) 0, and the degree of d(x)Is less than or equal to the degree of f(x), there exist unique polynomials q(x) and r(x) such that

f(x) = d(x)q(x) + r(x)

dividend quotient divisor remainder

Where r(x) = 0 or the degree of r(x) is less than the degree of d(x). If theRemainder r(x) is zero, d(x) divides evenly into f(x)

Before applying the Division Algorithm,1. Write the dividend and divisor in descending powers of the variable.2. Insert placeholders with zero coefficients for missing powers of the variable.


3 21 0 0 1x x x x 3 2x x

2 0x x

Example: Divide x3 +1 by x + 1.

2x x

1x

0

2x x 1

1x

Therefore, x3 + 1 = (x + 1)(x2 - x + 1)


Synthetic Division (use only when dividing by divisors of the form x – k)

This process requires only two mathematical operations:

Addition

Multiplication

To divide ax3 + b x2 + c x + d by x – k, use the following processStep 1: Set x – k equal to zero and solve for k. Place this number in the window.

Step 2: Beside the “window” write the coefficients of the variable in descending order along with the constant. Place a zero to hold the place of any missing power.

k a b c dwindow “the line”


__________

k a b c d

a

___ _______

k a b c d

ka

a

Synthetic Division (Cont.)

Step 3: Always bring down the first number in the line

Step 4: Multiply the number in the window by the number “below the line” and place under the next number “above the line”.

Step 5: Add ka and b and place the result below the line. Continue following Step 2until the end.

___ _______

k a b c d

ka

a


Example using synthetic division.

Divide x4 – 10x2 – 2x + 4 by x – 3.

3 1

1

3

3

9

-1

-3

-5

-15

-11

0 -10 -2 4

To write the quotient, remember that the initial power decreases by 1.

Here the quotient is x3 + 3x2 – x – 5 with a remainder of – 11.

Writing it in f(x) = (divisor)(quotient) + (remainder):

3 23 3 5 11f x x x x x


3 29 18 16 32 2x x x x

3 512 8x x

22 9 16f x x x

Try these using synthetic division.

Express your answer in f(x) = (divisor)(quotient) + (remainder) format.

1.

28 8 64f x x x x 2.


Two Important Theorems

The Remainder Theorem

If a polynomial f(x) is divided by x – k, the remainder isr = f(k)

This states that if f(x) is divided by x – k, the remainder will be the same as iff(x) was evaluated by f(k).

Example: Remember when x4 – 10x2 – 2x + 4 was divided by x – 3 usingsynthetic division, the remainder was – 11.

Evaluate f(x) = x4 – 10x2 – 2x + 4 by f(3). 4 2( ) 3 10 3 2 3 4 11f x

Notice that f(3) = - 11, the same as the remainder using synthetic division.


(1)f

( 2)f

(1/ 2)f

3( ) 4 13 10f x x x

Use Synthetic Division to find each of the following remainders. Then evaluate each function using the Remainder Theorem.

(8)f

1.

2.

3.

4.

f(1) = 1

f(-2) = 4

f(1/2) = 4

f(8) = 1978

Try these.


Two Important Theorems (cont.)

The Factor Theorem

A polynomial f(x) has a factor of x – k if and only if f(k) =0.

In the example x3 – 512 divided by x + 8, the remainder was zero.

Since f( – 8) equals zero, this tells one that x + 8 is a factor and the quotientx2 – 8x + 64 is also a factor.


The Rational Zero Test

If the polynomial

f(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0

Has integer coefficients, every rational zero of f has the form

Rational zero =

where p and q have no common factors other than 1, p is a factorof the constant term a0 and q is a factor of the leading coefficient an.

p

q


The Rational Zero Test (cont.)

What does this mean?

The Rational Zero Test gives one a list of all rational numbers which could be a zero of the given function.

How does one find them?

1. List the factors of both p and q.

2. Create fractions using p factors of constant

q factors of leading coefficient


3 22 3 8 3f x x x x

1, 3

1, 2

p

q

1 31, 3, ,

2 2

p

q

Find all possible rational zeros of

constant 3

leading coefficient 2

p

q


4 2( ) 4 5 6f x x x

3 210 15 16 12f x x x x

1 1 3 31, 2, 3, 6, , , ,

2 4 2 4

p

q

Try these.

Find all the possible rational zeros for the following:

1.

2.

1 1 1 2 3 3 3 4 6 121, 2, 3, 4, 6, 12, , , , , , , , , ,

2 5 10 5 2 5 10 5 5 5

p

q


Descartes Rule of Signs gives some insight on how many positive andnegative zeros exist for a polynomial function.

Descartes Rule of Signs

Let f(x) =anxn + an-1xn-1 + … + a2x2 + a1x + a0 be a polynomialwith real coefficients and a0 0.

1. 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 by an even integer.

2. The 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.


3 23 5 6 4f x x x x

Apply Descartes Rule of Signs to the following polynomial function to determine how many positive and negative real zeros may exist.

Count the number of sign changes in f(x) to determine the number ofpossible positive real zeros.

3 23 5 6 4f x x x x

+ +

1 2 3

Number of possible positive real zeros

Less an even number

3 or 1


3 23 5 6 4f x x x x

3 23 5 6 4f x x x x

3 23 5 6 4f x x x x

Applying Descartes Rule of Signs (cont.)

To find the number of possible negative real zeros, f(x) must be evaluatedfor f(-x) and then the sign changes counted.

3 23 5 6 4f x x x x Now count the sign changes.

0 0 0There are no sign changes; therefore, there are no negative zeros.


4 3 22 6 5f x x x x x

4 3 22 6 5f x x x x x

Apply Descartes Rule of Signs to the following problems to determinehow many possible positive and negative real zeros exists.

1.

Positive real zeros:

Negative real zeros:

3 24 2 3f x x x x

3 2( ) 4 2 3f x x x x 2.

Positive real zeros

Negative real zeros:

0

3 or 1

4, 2, or 0

0

Try these:


Upper and Lower BoundsThis is a test that can be applied while performing synthetic division to helprestrict the search for zeros of a function.

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

1. 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 f.

2. If 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.


1 6 4 3 2

6

5 3

2 5

6 2

Applying Upper and Lower Bounds

Example 1:

1 1 4 0 0 5

1 5 5 5

1 5 5 5 10

The last row shows all positives.One can conclude that 1 is an upper bound and no values greater than 1 will be a zero.

Example 2: The last row shows signs alternating from positive tonegative. One can conclude that –1 is a lower bound andno values less than –1 will be a zero.


3 22 3 12 8f x x x x

2f 4f

Try these:

Test to see if these are upper or lower bounds for the given function.

3f

4 2 3 12 8

8 20 32

2 5 8 40

2 2 3 12 8

4 2 20

2 1 10 12

3 2 3 12 8

6 27 45

2 9 15 37

Neither Upper Bound

Lower Bound

All positive

Signs alternate


3 26 4 3 2f x x x x Putting it all together!

Find all real zeros for

Step 1: Apply Descartes Rule of SignsPositive zeros: 3 or 1Negative zeros: 0

Step 2: Apply the Rational Zero Test

2 constant

6 leading coefficient

: 1, 2

: 1, 2, 3, 6

1 1 2 11, 2, , , ,

2 3 3 6

p

q

p factors

q factors

p

q

Note: Since there are no negativezeros, only the positive values need to be tried in synthetic division.


3 26 4 3 2f x x x x

1 6 4 3 2

6 2 5

6 2 5 3

2

6 4 3 23

4 0 2

6 0 3 0

Putting it all together! (cont.)


Step 3: Apply synthetic division to find a zero. Begin with integers and look for an upper bound. 2

2

2

6 3 0

6 3

1

2

2

2

x

x

x

ix

All positives

Upper Bound

Try values less than 1

2/3 is a zero because the remainder is 0.

The resulting line is nowrepresenting a quadraticfunction. Set this equal to zero and solve.

These are NOT realzeros.


3 26 4 3 2f x x x x

Putting it all together! (finishing it up)


Remember when applying the quadratic formula, 2 imaginary zeros were found.

These are NOT part of the solution because only real zeros were asked for.

Therefore, the only real zero for the function is 2

3


Wrapping it up!

When finding the real zeros of a function:

1. Apply Descartes Rule of Signs to determine how many positive andnegative real zeros may exist. Remember to always decrease by 2 as part of the rule.

2. Apply the Rational Zero Test to determine the rational possibilities forthe zeros.

3. Use synthetic division to find zeros. Look for upper and lower boundsto possibly eliminate some of the rational values given by the RationalZero Test.

4. Once you have found a zero, determine if the resulting line from synthetic division is a quadratic. If so, write the quadratic equation and solve. If not,continue with synthetic division always using the resulting line to find anotherzero. Try the exact same value in the window again to see if a double root exists. If not, try another possibility from the Rational Zero Test.


3 2 4 4f x x x x Try this.


Positive: 1Negative: 2 or 0

1, 2, 4p

q

Real Zeros are: -2, -1, and 2

Step 1: Descartes Rule of Signs

Step 2: Rational Zero Test

Step 3: Find the zeros using Synthetic Division and/or solve a quadratic equation


What you should know:

1. How to divide polynomials by other polynomials using long division.

2. How to divide polynomial by binominals in the form (x – k) usingsynthetic division and how to recognize upper and lower bounds.

3. How to apply the Remainder and Factor Theorems.

4. How to determine possible rational zeros of polynomial functions usingthe Rational Zero Test.

5. How to determine the number of positive and negative zeros a givenfunction has by applying Descartes Rule of Signs.

Documents

Section 2-3 Real Zeros of Polynomial Functions. Real Zeros of Polynomial Functions What you should know: 1. How to divide polynomials by other polynomials