§ 6.4 Division of Polynomials. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 6.4 Division...

§ 6.4

Division of Polynomials

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 6.4


In this section we will look at dividing by a monomial and dividing by a polynomial containing more than one term.

Division of a polynomial by a monomial is relatively easy – you just divide each term of the polynomial by the monomial. The number of separate divisions you will have is the number of terms in the polynomial.

The second case, that of dividing a polynomial by a polynomial having more than one term, is more difficult. That case requires a process of long division.



Dividing a Polynomial by a Monomial

To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.

1st case



EXAMPLEEXAMPLE

Divide: .10502040 3322334 yxyxyxyx

SOLUTIONSOLUTION

33

22334

10

502040

yx

yxyxyx

33

2

33

23

33

34

10

50

10

20

10

40

yx

yx

yx

yx

yx

yx

2

524

xyyx

Express the division in a vertical format.

Divide each term of the polynomial by the monomial. Note the 3 separate quotients.

Simplify each quotient.



Now, we will consider the harder problem – that ofdividing a polynomial by a polynomial having more than one term.

The four steps you use in long division of whole numbers – divide, multiply, subtract, bring down the next term – form the same repetitive procedure for polynomial long division.

Carefully consider and try to remember the four terms illustrated on the next slide. These terms are: quotient, divisor, dividend, and remainder.



4

1

84

742

x

xx

DIVISOR QUOTIENT

EXAMPLEEXAMPLE

REMAINDER

DIVIDEND

2nd case



EXAMPLEEXAMPLE

Divide: . 2275 23 xxxx

SOLUTIONSOLUTIONArrange the terms of the dividend, , and the divisor, (x + 2), in descending powers of x.

Divide (the first term in the dividend) by x (the first term in the divisor). Align like terms.

2752 23 xxxx 275 23 xxx

2

23 2752

x

xxxx 3x



Multiply each term in the divisor (x + 2) by , aligning terms of the product under like terms in the dividend.

Subtract from by changing the sign of each term in the lower expression and then adding.

2

23

23

2

2752

x

xx

xxxx

2x

CONTINUECONTINUEDD

2

2

23

23

3

2

2752

x

x

xx

xxxx

23 2xx 23 5xx



Bring down 7x from the original dividend and add algebraically to form a new dividend.

Find the second term of the quotient. Divide the first term of by x, the first term of the divisor.

CONTINUECONTINUEDD 2

2

23

23

73

2

2752

x

xx

xx

xxxx

xx 73 2

xx

xx

xx

xxxx

3

73

2

2752

2

2

23

23



Multiply the divisor (x + 2) by 3x, aligning under like terms in the new dividend. Then subtract.

CONTINUECONTINUEDD

xx

x

xx

xx

xx

xxxx

3

63

73

2

2752

2

2

2

23

23



Bring down 2 from the original dividend and add algebraically to form a new dividend.

Find the third term of the quotient, 1. Divide the first term of x + 2 by x, the first term of the divisor.

Multiply the divisor by 1, aligning under like terms in the new dividend. Then subtract to obtain the remainder of 0.

CONTINUECONTINUEDD 13

0

2

2

63

73

2

2752

2

2

2

23

23

xx

x

x

xx

xx

xx

xxxx



The quotient is and the remainder is 0. We will not list a remainder of 0 in the answer. Thus,

CONTINUECONTINUEDD

132 xx

.132275 223 xxxxxx


Long Division of Polynomials

Long Division of Polynomials1) Arrange the terms of both the dividend and the divisor in descending powers of any variable.

2) Divide the first term in the dividend by the first term in the divisor. The result is the first term of the quotient.

3) Multiply every term in the divisor by the first term in the quotient. Write the resulting product beneath the dividend with like terms lined up.

4) Subtract the product from the dividend.

5) Bring down the next term in the original dividend and write it next to the remainder to form a new dividend.

6) Use this new expression as the dividend and repeat this process until the remainder can no longer be divided. This will occur when the degree of the remainder (the highest exponent on a variable in the remainder) is less than the degree of the divisor.



EXAMPLEEXAMPLE

Divide: . 281243 2235 xxxxx

SOLUTIONSOLUTION

We write the dividend, , as to keep all like terms aligned. For the same reason, we write the divisor, , as

81243 235 xxxx

22 x812403 2345 xxxxx

.202 xx



453

2

804

824

1005

1245

603

81240320

3

2

2

23

23

345

23452

xx

x

xx

xx

xxx

xxx

xxx

xxxxxxx

CONTINUECONTINUEDD



CONTINUECONTINUEDD

The division process is finished because the degree of -2x, which is 1, is less than the degree of the divisor , which is 2. The answer is

22 x

.x

xxx

x

xxxx

2

2453

2

812432

32

235



Important to Important to Remember:Remember:

To divide by a polynomial containing more than one term, use long division. If necessary, arrange the dividend in descending powers of the variable. Do the same with the divisor. If a power of a variable is missing in the dividend, add that term using a coefficient of 0.

Repeat the four steps of the long-division process – divide, multiply, subtract, bring down the next term – until the degree of the remainder is less than the degree of the divisor.

When the degree of the remainder is less than the degree of the divisor – you know you are done!

§ 6.4 Division of Polynomials. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 6.4 Division...

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