Warm up

Lesson 4-3 The Remainder and Factor Theorems

Objective: To use the remainder theorem in dividing polynomials

Synthetic DivisionTo divide a polynomial by x – r

Example1. 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).

The Remainder Theorem

• If the polynomial f (x) is divided by x – r, then the remainder is f (r).– Example using synthetic division on the

first warm up 3x2 – 11x + 5 we got a remainder of 9

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

The Remainder Theorem

• You can use synthetic division to check the remainder theorem.

• x-3• f(3) = 2(3)3 – 3(3)2 – 2(3) + 1 • = 54 – 27- 6 + 1 • = 22 therefore R = 22 now do the

synthetic division: 3 2 -3 -2 +1

• The last number is 22

1232)( 23 xxxxf

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