Divide 4 – 5x – x2 + 6x3 by 3x – 2.Solution Write the divisor and dividend in descending powers of x. Next, consider how many times 3x divides into 6x3.

3x – 2 6x3 – x2 – 5x + 42x2 Divide: 6x3/3x = 2x2

6x3 – 4x2Multiply

Multiply: 2x2(3x – 2)

3x2 – 2x Subtract 3x2 – 2x from 3x2 – 5x-3x

3x2

Subtract 6x3 – 4x2 from 6x3 – x2

Date:2.4 Real Zeros of Polynomial Functions

– 5xand bring down –5x.

Divide: 3x2/3x

+ x

+ 4 and bring down 4.

Divide: -3x/3x

– 1

Multiply -1(3x – 2) -3x + 2

Subtract -3x + 2 from -3x + 4, 2 leaving a remainder of 2.

or +2

3x 2

Multiplyx(3x – 2)

The Division AlgorithmIf f (x) and d(x) are polynomials, with d(x) = 0, and the degree of d(x) is less than or equal to the degree of f (x), then there exist unique polynomials q(x) and r(x) such that

f (x) = d(x) • q(x) + r(x).

 The remainder, r(x), equals 0 or its is of degree less than the degree of d(x).

If r(x) = 0, we say that d(x) divides evenly in to f (x) and that d(x) and q(x) are factors of f (x).

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

f (x) = d(x) • q(x) + r(x).

 The remainder, r(x), equals 0 or its is of degree less than the degree of d(x).

If r(x) = 0, we say that d(x) divides evenly in to f (x) and that d(x) and q(x) are factors of f (x).

Dividend Divisor Quotient Remainder

33

34232

23

xx

xxx

Example

Divide:

342333 232 xxxxx

Example cont.

3x

342333 232 xxxxx

Example cont.

3x

3x3 + 9x2 + 9x342333 232 xxxxx

Example cont.

3x

342333 232 xxxxx

Example cont.

-(3x3 + 9x2 + 9x)

-11x2 - 5x - 3

3x -11

-(3x3 + 9x2 + 9x)

-11x2 - 5x - 3

342333 232 xxxxx

Example cont.

3x -11

-(3x3 + 9x2 + 9x)

-11x2 - 5x - 3

-11x2 - 33x - 33

342333 232 xxxxx

Example cont.

3x -11

-(3x3 + 9x2 + 9x)

-11x2 - 5x - 3

-(-11x2 - 33x - 33)

28x+30

342333 232 xxxxx

Example cont.

3x -11

-(3x3 + 9x2 + 9x)

-11x2 - 5x - 3

-(-11x2 - 33x - 33)

28x+30

342333 232 xxxxx

Example cont.

28x 30x2 3x 3

x2 14x 45 by x 9Complete Student Checkpoint

Divide

x2 14x 45 by x 9Complete Student Checkpoint

Divide

x 9 x2 14x 45

x

x2 + 9x

x2 14x 45 by x 9Complete Student Checkpoint

Divide

x 9 x2 14x 45

x

-(x2 + 9x)

x2 14x 45 by x 9Complete Student Checkpoint

Divide

x 9 x2 14x 45

x

-(x2 + 9x)

5x + 45

x2 14x 45 by x 9Complete Student Checkpoint

Divide

x 9 x2 14x 45

x + 5

-(x2 + 9x)

5x + 45

5x + 45

x2 14x 45 by x 9Complete Student Checkpoint

Divide

x 9 x2 14x 45

x + 5

-(x2 + 9x)

5x + 45

-(5x + 45)

x2 14x 45 by x 9Complete Student Checkpoint

Divide

x 9 x2 14x 45

x + 5

-(x2 + 9x)

5x + 45

-(5x + 45)

0

x2 14x 45 by x 9Complete Student Checkpoint

Divide

x 9 x2 14x 45

x + 5

-(x2 + 9x)

5x + 45

-(5x + 45)

0

x2 14x 45 by x 9Complete Student Checkpoint

Divide

x 9 x2 14x 45

Synthetic Division

1. Arrange polynomials in descending powers, with a 0 coefficient for any missing terms.

To divide a polynomial by x – c

2. Write k for the divisor, x – k. To the right, write the coefficients of the dividend.

x 3 x3 4x2 5x 5

3. Write the leading coefficient of the dividend on the bottom row.

3 1 4 -5 5

1

4. Multiply k (in this case, 3) times the value just written on the bottom row. Write the product in the next column in the 2nd row.

Multiply by 3

3

5. Add the values in this new column, writing the sum in the bottom row.

7

Add

6. Repeat this series of multiplications and additions until all columns are filled in. Add.

1x2 + 7x + 16 +53

x – 3

Synthetic Division

4853

3 1 4 -5 5

3

1 7

7. Use the numbers in the last row to write the quotient and remainder in fractional form. The degree of the first term of the quotient is one less than the degree of the first term of the dividend. The final value in the row is the remainder.

16Multiply by 3.21

Text Example

Use synthetic division to divide 5x3 + 6x + 8 by x + 2. The divisor must be in the form x – k. Write x + 2 as x – (-2). This means that k = -2. Write a 0 coefficient for the missing x2-term in the dividend.

-2 5 0 6 8

Use the coefficients of the dividend in descending powers of x.

This is c in x-(-2).

5

-10

-10

20

26

-52

-44

Thus,x + 2 5x3 + 0x2 + 6x + 8

5x2 – 10x + 26 – x + 244

Complete Student Checkpoint

Use synthetic division to divide x3 7x 6 by x 2

1

-2

-2

4

-3

6

0

Thus,x + 2 x3 + 0x2 - 7x - 6

1x2 – 2x – 3

-2 1 0 -7 -6

The Remainder TheoremIf the polynomial f(x) is divided by x – k, then

the remainder is f(k).

The Factor TheoremLet f(x) be a polynomial.If f(k) = 0, then x – k is a factor of f(x).If x – k is a factor of f(x), then f(k) = 0.

Text Example

Find f(-2) for f(x)= 5x3 + 6x + 8

f(-2)= 5(-2) 3 + 6(-2) + 8

= 5(-8) + -12 + 8

= -40 + -12 + 8

= -52 + 8

= -44

-2 5 0 6 8

5

-10

-10

20

26

-52

-44

Here is the previous problem. Notice the remainder:

Synthetic divisions remainder is an alternative to plugging in a x-value into the function. It is referred to in the remainder theorem.

Text ExampleSolve the equation 2x3 – 3x2 – 11x + 6 = 0 given that 3 is a zero of f(x) = 2x3 – 3x2 – 11x + 6.

Given that f (3) = 0. The Factor Theorem tells us that x – 3 is a factor of f(x). Use synthetic division to divide f (x) by x – 3.

Equivalently, 2x3 – 3x2 – 11x + 6 = (x – 3) (2x2 + 3x – 2)

3 2 -3 -11 6

2x2 + 3x – 22x3 – 3x2 – 11x + 6x – 3

2

6

3

9

-2

-6

0

The solution set is {-2, 1/2 , 3}.

Now we can solve the polynomial equation.

2x3 – 3x2 – 11x + 6 = 0 This is the given equation.

(x – 3) (2x2 + 3x – 2) = 0 Factor using the result from the synthetic division.

(x – 3)(2x – 1)(x + 2) = 0 Factor the trinomial.

x – 3 = 0 or 2x – 1 = 0 or x + 2 = 0Set each factor equal to 0.

x = 3 x = 1/2 x = -2 Solve for x.

Text Example cont.

Equivalently, 2x3 – 3x2 – 11x + 6 = (x – 3) (2x2 + 3x – 2)

ExampleFind a third-degree polynomial function f(x) with real coefficients that has -4, 3 and 5 as zeros and such that f(0)=180

f (x) an(x )(x )(x ) ( 4) 3 5

f (x) an(x 4)(x 3)(x 5)

f (0) an(0 4)(0 3)(0 5)

180 an(4)( 3)( 5)

180 60an

60 60

3 an

f (x) 3(x2 x 12)(x 5)

f (x) (3x2 3x 12)(x 5)

f (x) 3x3 3x2 12x 15x2 15x 60

f (x) 3x3 12x2 27x 60

Complete Student CheckpointGiven f(x)=3x3 4x2 5x 3, use

the Remainder Theorem to find f(-4)

3

-12

-8

32

27

-108

-105

-4 3 4 -5 3

f (-4) -105

Complete Student CheckpointSolve the equation 15x3 14x2 3x 2 0 given that

-1 is a zero of f (x) 15x3 14x2 3x 2

15

-15

-1

1

-2

2

0

-1 15 14 -3 -2

15x3 14x2 3x 2 0

x 1 15x2 x 2 0

X

+

5-6

-1

-30

15x2

-2 5x

-6x3x

5x – 2

1x 1 5x 2 3x 1 0

x 1 , 2

5 , -

1

3

Upper and Lower Bound Tests for Real ZerosIf f (x) anx

n an1xn1 … a2x

2 a1x a0 be a polynomial of degree n > 1 with a positive leading coefficient. Suppose f (x) is divided by x - k using synthetic division:

1. If k > 0 and every number in the last line is non negative (positive or zero), then k is an upper bound for the real zeros of f.

2. If k > 0 and the numbers in the last line are alternating nonnegative and nonpositive, then k is a lower bound for the real zeros of f.

DAY 2

ExampleProve that all of the zeros must lie in the interval [-2, 5]for: f (x) 2x4 7x3 8x2 14x 8

Prove that 5 is an upper bound (use synthetic division):

since the last line is all positive, 5 is an upper bound

2 -7 -8 14 8 5 10 15 35 -245 2 3 7 49 253

2 -7 -8 14 8-2 -4 22 -28 28 2 -11 14 -14 36

Prove that -2 is a lower bound:

since the last line alternates signs, -2 is a lower bound

Graph f(x) on a graphing calculator to view these upper and lower bounds for the zeros of the function. Use window [-2,5] by [-50,50].

The Rational Zero Theorem

If f (x) anxn an-1x

n-1 … a1x a0 has integer

coefficients and

p/q is a rational zero(or factor of f(x)), then

p is a factor of the constant term a0 and

q is a factor of the leading coefficient an.

Example

Find all of the possible real, rational roots of f(x) = 2x3-3x2+5

p is a factor of 5 = +1, +5

q is a factor of 2 = +1, +2

Roots are: p/q = +1, +1/2, +5, +5/2

ExampleFind all zeros of f(x) = x3+12x2+21x+10

p/q = +1, +2, +5, +10 are possible zerosStart plugging them in to find the zeros or use synthetic

division:

f(1) = 44

f(-1) = 0 -1 is a zero

Since f(-1)=0, x+1 is a factor, or a zero. Divide out -1 to get the other factor. Since we did synthetic division, we have already done that so: x3+12x2+21x+10 = 0

1 12 21 10 1 1 13 34 1 13 34 44-1 -1 -11 -10 1 11 10 0

(x + 1)(x2+11x+10) = 0(x + 1)(x + 1)(x+10) = 0

The solutions are -1, and -10

Complete Student CheckpointList all possible rational zeros of f (x) 4x5 12x4 4x 3

p is a factor of 3 = +1, +3

q is a factor of 4 = +1, +2 , +4

Roots are: p/q = +1, +1/2, +1/4, +3, +3/2, +3/4

Real Zeros of Polynomial Functions