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O B J E C T I V E S :

1. D E F I N E A P O LY N O M I A L .2. D I V I D E P O LY N O M I A L S .3. A P P LY T H E R E M A I N D E R A N D FA C T O R T H E O R E M S ,

A N D T H E C O N N E C T I O N S B E T W E E N R E M A I N D E R S A N D FA C T O R S .

4. D E T E R M I N E T H E M A X I M U M N U M B E R O F Z E R O S O F A P O LY N O M I A L .

4.1 Polynomial Functions

Polynomials

Characteristics of Polynomials:A sum or difference of monomialsAll exponents are whole numbersNo variable contained in the denominatorNo variable is under a radical

Polynomials Not Polynomialsx3 6x2 1

2y 15 y10 7

w 6.7

12

x 2 5

2y13 7y 1

m2 m

p 2 5p 3

Degree of a Polynomial

Example #1Polynomial Division

Divide: 2x4 25x2 30x 5 by x 4Expand the dividend and fill in missing terms, then follow the same steps as when dividing real numbers: divide, multiply, subtract, bring down.

53025024 234 xxxxx2x3

8x3

34

22 xxx

Divide

34

3

82

42

xx

xx

Multiply(2x4 − 8x3)

Subtract

Bring down

– 25x2

Once this cycle is made once, it is repeated until the degree of the remainder is less than the degree of the dividend.


Divide: 2x4 25x2 30x 5 by x 4

53025024 234 xxxxx2x3

8x3 23

88 xxx

Divide

23

2

328

48

xx

xx

Multiply

(2x4 − 8x3)

Subtract

Bringdown– 25x2

+ 8x2

(8x3 – 32x2)

7x2 – 30x



53025024 234 xxxxx2x3

8x3x

xx 77 2

Divide

xx

xx

287

472

Multiply

(2x4 − 8x3)

Subtract

Bringdown– 25x2

+ 8x2

(8x3 – 32x2)

7x2 – 30x

+ 7x

(7x2 – 28x)

−2x + 5



53025024 234 xxxxx2x3

8x322

xx

Divide

8242

xx

Multiply

(2x4 − 8x3)

Subtract

– 25x2

+ 8x2

(8x3 – 32x2)

7x2 – 30x

+ 7x

(7x2 – 28x)

−2x + 5

− 2

(−2x + 8)−3 Remainder

The degree of the remainder is less than that of the divisor so the cycle is completed.



53025024 234 xxxxx2x3

8x3

43

x

(2x4 − 8x3)

– 25x2

+ 8x2

(8x3 – 32x2)

7x2 – 30x

+ 7x

(7x2 – 28x)

−2x + 5

− 2

(−2x + 8)−3

Finally the remainder is put over the divisor and added to the quotient.

Example #2Synthetic Division

Divide using synthetic division:x 4 5x 3 2x 2 6x 17 by x 5

Caution! Synthetic division only works when the divisor is a first-degree polynomial of the form (x − a).

505

xx

−5 1 5 2 6 −17Set the divisor equal to 0 and solve. Place it in the left corner in a half-box

Write the leading coefficients of the dividend in order of degree next to the half-box. Include 0’s for missing terms if necessary.

1Bring down the first digit.

−5

Add the column and repeat the process.

0

Multiply it by the -5.Write the answer under the next digit.



−5 1 5 2 6 −17

1

−5


0

Multiply the sum again by the -5.

Write the answer under the next digit.

0

2



−5 1 5 2 6 −17

1

−5


0



0

2 −4

−10



−5 1 5 2 6 −17

1

−5

Add the column. The last digit forms the remainder.

0



0

2 −4

−10

3

20



−5 1 5 2 6 −17

1

−5

0

0

2 −4

−10

3

20

After using up all terms, the quotient needs the variables added back into the final answer. The first term is always one degree less than the dividend. Remember to place the remainder over the original divisor.

5342

534201

3

23

xxx

xxxx

Example #3Factors Determined by Division

Determine if 3x 2 2 is a factor of 3x 3 15x 2 2x 10.

102153203 232 xxxxx

xxx

2

3

33

xxx

xxx

203

20323

2

x

(3x3 + 0x2 – 2x)Divide

Multiply

Subtract 15x2 + 0x −10

Bring down



102153203 232 xxxxx

53

152

2

xx

10015

20352

2

xx

xx

x

(3x3 + 0x2 – 2x)Divide

MultiplySubtract

15x2 + 0x −10

+ 5

(15x2 + 0x – 10)0



102153203 232 xxxxxx

(3x3 + 0x2 – 2x)15x2 + 0x −10

+ 5

(15x2 + 0x – 10)0

Because the remainder is 0, the divisor divides evenly into the dividend and is then said to be a factor of the dividend.

Remainder Theorem

If a polynomial f(x) is divided by x – c, then the remainder is f(c).

Example #4The remainder when dividing by x – c:

Find the remainder when x 87 2x36 8 is divided by x 1.

Using the remainder theorem, this problem can be performed without actually dividing the dividend by the divisor.

101

xx

7821

81211 3687

f

First find the value of c in the divisor x – c.

Next evaluate f(c).

The remainder is therefore 7.

Example #5The remainder when dividing by x + k

303

xx

33615108162

61527481263534323 34

f

Find the remainder when 2x4 + 4x3 + 5x – 6 is divided by x + 3.

Factor Theorem

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

The idea in this theorem was actually introduced in Example #3.

The difference is that this theorem only works for linear factors, in which case division is not necessary,

otherwise you must use division to check if they are factors.

Example #6The Factor Theorem

Show that x 2 is a factor of x 3 5x2 2x 24 by using the Factor Theorem.

Find q(x) such that (x 2)q(x) x 3 5x 2 2x 24.

202

xx

0244208

24222522 23

f

Since f (2) = 0, then it is a factor of the polynomial. Next we’ll find the actual quotient using synthetic division.

2 1 5 −2 −24

1 7 12 0

2 14 24

1272 xx

The answer can be checked by multiplication.

2425

24142127

1272

23

223

2

xxx

xxxxx

xxx

Fundamental Polynomial Connections

If one of the following are true, all are true:

A. r is a zero of the function fB. r is an x-intercept of the graph of the

function fC. x = r is a solution, or root, of the function

f(x) = 0D. x – r is a factor of the polynomial f(x)

Example #7Fundamental Polynomial Connections

For f(x) 2x 3 7x 2 20x 25, find the following:

a) the x-intercepts of the graph of f

b) the zeros of f

c) the solutions to 2x3 7x 2 20x 25 0

d) the linear factors with realcoefficients of 2x 3 7x 2 20x 25

1 2 3 4 5–1–2–3–4–5 x

10

20

30

40

50

–10

–20

–30

–40

–50

y

From the graph it is clear that the x-intercepts are at −2.5, 1, & 5.The zeros of f are also −2.5, 1, & 5.The solutions are x = −2.5, x = 1, & x = 5.Setting each solution equal to 0 we obtain the linear factors of (2x + 5), (x – 1), & (x – 5).

052

02252

025

05.2

x

x

x

x

**Note**:

Example #8A Polynomial with Specific Zeros

Find three polynomials of different degrees that have

-2, 1, and 3 as zeros. 3125ofdegree

3124ofdegree

3123ofdegree

22

2

xxxxk

xxxxh

xxxxg

The simplest polynomial must be of degree 3 to have all the zeros requested. For a degree of 4, one (not necessarily the factor shown) of the factors must repeat twice, but highest exponent (if it were multiplied out) would still only be 4. Finally, for a degree of 5, two of the factors must repeat.

Example #8A Polynomial with Specific Zeros

Find three polynomials of different degrees that have

-2, 1, and 3 as zeros. 131275ofdegree

3125ofdegree2

22

xxxxxp

xxxxk

Here, an alternative polynomial of degree 5 is shown that retains the same zeros as the one above it. First of all, the 7 out front is a constant and has NO effect on where the graph crosses the x-axis. This could be any real number, even a negative.

Secondly, the final group of (x2 + 1) has imaginary roots at ±i. When everything is multiplied out, the resulting polynomial will still have a degree of 5, but the imaginary roots will have no effect on where the graph crosses the x-axis either.

Number of Zeros

A polynomial of degree n has at most n distinct real zeros.

Example #8Determining the Number of Zeros

Find the maximum number of zeros that the polynomial could contain.

The degree of this polynomial is 5.

At most, this polynomial can have only 5 distinct real zeros.

This means that at most, this polynomial will cross the x-axis 5 times.

726)( 35 xxxxf

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