Beginning Algebra, 11th Edition

Step 7 CHECK

Multiply Divisor (Quotient

including the Remainder).

= 4 x^3 - 4 x^2 + 5 x- 8 ✓ NOW TRY

= 4 x^3 - 2 x^2 - 2 x^2 + x+ 4 x- 2 - 6

= 12 x- 1212 x^22 + 12 x- 121 - x 2 + 12 x- 12122 + 12 x- 1 2a

- 6

2 x- 1

b

*

12 x- 1 2a 2 x^2 - x+ 2 +

- 6

2 x- 1

b

NOW TRY
EXERCISE 6
Divide.

6 k^3 - 20 k-k^2 + 1

2 k- 3

SECTION 5.7 Dividing Polynomials 345

Dividing into a Polynomial with Missing Terms

Divide by

Here, the dividend, is missing the -term and the x-term. We use 0 as the

coefficient for each missing term. Thus,

0

x- 1

x- 1

x^2 - x

x^2 + 0 x

x^3 - x^2

x- 1 x^3 + 0 x^2 + 0 x- 1

x^2 + x+ 1

x^3 - 1 = x^3 + 0 x^2 + 0 x-1.

x^3 - 1, x^2

x^3 - 1 x-1.

EXAMPLE 7

Insert placeholders

for the missing terms.

The remainder is 0. The quotient is

CHECK ✓

NOW TRY

Dividing by a Polynomial with Missing Terms

Divide by

Since the divisor, has a missing x-term, write it as

- 3 x- 2 Remainder

x^2 + 0 x+ 1

x^2 - 3 x- 1

2 x^3 + 0 x^2 + 2 x

2 x^3 + x^2 - x

x^4 + 0 x^3 + x^2

x^2 + 0 x+ 1 x^4 + 2 x^3 + 2 x^2 - x- 1

x^2 + 2 x+ 1

x^2 +1, x^2 + 0 x+1.

x^4 + 2 x^3 + 2 x^2 - x- 1 x^2 + 1.

EXAMPLE 8

1 x- 121 x^2 + x+ 12 = x^3 - 1 Divisor*Quotient=Dividend

x^2 +x+1.

Insert a

placeholder for the

missing term.

When the result of subtracting ( here) is a constant or a polynomial of degree

less than the divisor that constant or polynomial is the remainder.

The answer is

x^2 + 2 x+ 1 +

- 3 x- 2

x^2 + 1

.

1 x^2 + 0 x+ 12 ,

- 3 x- 2

Remember to write

””+^ remainderdivisor^.

Multiplyto check that this is correct. NOW TRY

CAUTION Remember to include “+ ” as part of the answer.

remainder

divisor^

NOW TRY
EXERCISE 7
Divide by
m-10.

m^3 - 1000

NOW TRY ANSWERS

6. 
   

7. m^2 + 10 m+ 100

3 k^2 + 4 k- 4 +

• 11
  2 k- 3

NOW TRY
EXERCISE 8
Divide.

by y^2 +2.

y^4 - 5 y^3 + 6 y^2 +y- 4

8. y^2 - 5 y+ 4 +

11 y- 12

y^2 + 2