Warm up Divide using polynomial long division:

n2 – 9n – 22 n+2

Lesson 5-4 Synthetic Division

Objective: To use synthetic division to divide polynomials.

Synthetic Division Used as a shortcut when dividing a polynomial

by a first degree binomial.

It is very similar to polynomial division, only using the coefficients and not the variables.

Synthetic Division

To divide a polynomial by x – rExample

1. Arrange polynomials in descending powers, with a 0 coefficient for any missing terms. x – 3 x3 + 4x2 – 5x + 5

2. Write r for the divisor, x – r. To the right, 3 1 4 -5 5

write the coefficients of the dividend.

3. Write the leading coefficient of the dividend 3 1 4 -5 5

on the bottom row. Bring down 1.

1

4. Multiply r (in this case, 3) times the value 3 1 4 -5 5

just written on the bottom row. Write the 3 product in the next column in the 2nd row. 1

To divide a polynomial by x – rExample

1. Arrange polynomials in descending powers, with a 0 coefficient for any missing terms. x – 3 x3 + 4x2 – 5x + 5

2. Write r for the divisor, x – r. To the right, 3 1 4 -5 5

write the coefficients of the dividend.

3. Write the leading coefficient of the dividend 3 1 4 -5 5

on the bottom row. Bring down 1.

1

4. Multiply r (in this case, 3) times the value 3 1 4 -5 5

just written on the bottom row. Write the 3 product in the next column in the 2nd row. 1

Multiply by 3.

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

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

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.

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

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

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.

3

1

4

-5

5

3

1

7

Add.

3

1

4

-5

5

3

21

48

1

7

16

53

Add.

Multiply by 3.

3

1

4

-5

5

3

21

1

7

16

Add.

Multiply by 3.

1x2 + 7x + 16 +53x – 3

Synthetic Division

Example

Use synthetic division to divide 5x3 + 6x + 8 by x + 2.

Solution The divisor must be in the form x – r. Thus, we write x + 2 as x – (-2). This means that c = -2. Writing a 0 coefficient for the missing x2-term in the dividend, we can express the division as follows:

x – (-2) 5x3 + 0x2 + 6x + 8 .

Now we are ready to set up the problem so that we can use synthetic division.

-2

5

0

6

8

Use the coefficients of the dividend in descending powers of x.This is r in x-(-2).

Solution We begin the synthetic division process by bringing down 5. This is following by a series of multiplications and additions.

5. Add: 6 + 20 = 26.

-2

5

0

6

8

-10

20

5

-10

26

Add.

7. Add: 8 + (-52) = -44.

-2

5

0

6

8

-10

20

-52

5

-10

26

-44

Add.

4. Multiply: -2(-10) = 20.

-2

5

0

6

8

-10

20

5

-10

6. Multiply: -2(26) = -52.

-2

5

0

6

8

-10

20

-52

5

-10

26

-2

5

0

6

8

5

1. Bring down 5. 3. Add: 0 + (-10) = -10.

-2

5

0

6

8

-10

5

-10

Add.

2. Multiply: -2(5) = -10.

-2

5

0

6

8

-10

5

Example cont.

Solution

-2

5

0

6

8

-10

20

-52

5

-10

26

-44

The numbers in the last row represent the coefficients of the quotient and the remainder. The degree of the first term of the quotient is one less than that of the dividend. Because the degree of the dividend is 3, the degree of the quotient is 2. This means that the 5 in the last row represents 5x2.

Thus,

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

5x2 – 10x + 26 – x + 2

44

Example cont.

Practice

3

286 23

x

xxx

3

1132

x

xxAnswer:

The Remainder Theorem

If the polynomial f (x) is divided by x – r, then the remainder is f (r). Example using synthetic division 3x2 – 11x + 5 we got a remainder

of 9 x – 4 f (4) = 3(4)2 – 11(4) + 5 = 48 -44 +

5 =9

Practice x+2623)( 2 xxxf

Factor Theorem

P(x) has a factor x –r if and only if P(r)=0.

Example: Is x + 2 a factor of

? Use the remainder theorem to see if P(-

2) = 0 P(-2) = (-2)3 –(-2)2 – 2(-2) +8 = -8 – 4 + 4 + 8 = 0 therefore x + 2 is a factor.

82)( 23 xxxxf

Practice

Is x – 1 a factor of 126433)( 23456 xxxxxxxf