Divide Polynomials

Algebra Help: Section 2.22

Learn to divide polynomials by other polynomials, with algebra.

The terms of both divisor and dividend must be arranged in the order of the powers of the letters. Since the dividend is always the product of the divisor by the quotient sought, the highest power of any letter of the dividend must be the product of the highest powers of the same letter in the divisor and quotient. The inferior powers of this letter in the dividend must be the products of inferior powers in divisor and quotient.

See Simplified Steps below, for a detailed outline, explaining how to solve this problem and other similar polynomial divisions.

1. Divide
2a⁴ + 5a²b² + 2a³b - 6ab³ + 4b⁴
by
a² + 2ab + 4b².

1st Dividend 
2a⁴ + 2a³b + 5a²b²  - 6ab³ + 4b⁴
2a⁴ + 4a³b + 8a²b²
_______________________________
1st Rem or 2nd Dividend
-2a³b - 3a²b² - 6ab³ + 4b⁴
-2a³b - 4a³b² - 8ab³ + 4b⁴
_______________________________
2nd Rem or 3rd Dividend
a²b² + 2ab³ + 4b⁴ 
a²b² + 2ab³ + 4b⁴
        

Divisor   a² + 2ab  + 4b²
Quotient  2a² - 2ab + b²

ANALYSIS - Arrange the terms of both divisor and dividend according to the powers of a, so that in the dividend the exponents of this letter, taken in their order are 4,3,2,1. The order of the exponents of the divisor, in this letter, taken in their order are 2, 1. Now, according to the principle just stated, the first term of the dividend, thus arranged, must be equal to the first term of the divisor multiplied by that term of the quotient having the highest power of a.

Divide 2a⁴, the first term of the dividend by a², the first term of the divisor, and obtain 2a², for the first quotient.

We next multiply the whole divisor by this term of the quotient, and subtract the product from the dividend, bringing down as many terms as are necessary for a new dividend.

We then divide -2a³b, the first term of the remainder, by a², the first term of the divisor, and obtain -2ab for the second term of the quotient.

We next multiply the whole divisor by this term of the quotient, and subtract the product from the second dividend, and obtain a second remainder, to which we annex another term of the dividend for another dividend.

Dividing a³b³, the first term of this dividend, by a² the first term of the divisor, we obtain b², the third term of the quotient.

Last multiply the whole divisor by the third term of the quotient. Subtract the product from the last dividend. We have no remainder and the work is finished.

Rules

Arrange both divisor and dividend in order of the powers of one of the letters.
Divide the first term of the dividend by the first term of the divisor. The result equals the first term of the quotient.
Multiply the entire divisor by the first quotient. Subtract the product from the dividend.
Arrange the remainder for a new dividend, with which to proceed as before, till the first term of the divisor is no longer contained in the first term of the remainder.
Write the final remainder, if there be any, over the divisor in the form of a fraction, and the entire result will be the quotient sought.

Simplified Steps

The following steps solve the example problem, with variables representing dividends, the first term in a dividend, products, quotients, the divisor and first term in the divisor. This section was not included in the original book. With ease the pseudo code presented here could apply, as an algorithm, to solve polynomial division with code.

Loop through steps in similar division problems. Divide the current dividend's first term by the divisor's first term. The result's the first term in the quotient. Multiply the divisor by the first term in the quotient for the first product. Subtract the first product from the dividend for the second dividend. Repeat the process until a dividend's first term does not contain the first term of the divisor.

Each loop creates a quotient Q[n], product Product<n> and dividend Dividend<n>. The first term in the dividend's represented as Dividend<n>[1], where <n> is replaced with a digit such as Dividend1[1] or Dividend2[1]. Each step also demonstrates how to apply the process with the example polynomials.

The entire divisor's represented as Divisor. The first term of the divisor's represented as Divisor[1]. Both the entire divisor and the first term of the divisor, remain the same and are used in each loop. In this example the entire divisor is, a² + 2ab + 4b². The divisor's first term is a².

Start with the following dividend and divisor.

Dividend 2a⁴ + 2a³b + 5a²b²  - 6ab³ + 4b⁴
Divisor   a² + 2ab + 4b²

Divide the dividend's first term by the divisor's first term. 2a⁴ ÷ a² = 2a², which is the first quotient.
Dividend1[1]/Divisor[1] = Q[1]
Multiply the divisor by the first term in the quotient. The result's the first product.
Divisor * Q[1] = P1
```
 a² + 2ab + 4b² *
2a²
__________________
2a⁴ + 4a³b + 8a²b²
```

Subtract the first product from the original first dividend. The difference is the second dividend.
Dividend1 - P1 = Dividend2

2a⁴ + 2a³b + 5a²b²  - 6ab³ + 4b⁴
2a⁴ + 4a³b + 8a²b²
_______________________________
     -2a³b - 3a²b² - 6ab³ + 4b⁴

Divide the second dividend's first term, by the divisor's first term. The result is the second term of the quotient.
Dividend2[1]/Divisor[1] = Q[2]
```
-2a³b divide by
  a²
_____
-2ab
```
Multiply the divisor by the second term of the quotient. The result's our second product.
Divisor * Q[2] = P2
```
 a² + 2ab + 4b² *
-2ab
____________________
-2a³b - 4a²b² - 8ab³
```
Subtract the second product from the second dividend. The difference is the third dividend.
Dividend2 - P2 = Dividend3
```
-2a³b - 3a²b² - 6ab³ + 4b⁴ minus
-2a³b - 4a²b² - 8ab³
__________________________
         a²b² + 2ab³ + 4b⁴
```
Divide the first term of the third dividend by the first term of the divisor. The result's the third term of the quotient.
Dividend3[1]/Divisor[1] = Q[3]
```
a²b² divide by
a²
_____
  b²
```
Multiply the divisor times the third term of the quotient for the third product.
Divisor * Q[3] = P3
```
a² + 2ab + 4b² *
b²
_________________
a²b² + 2ab³ + 4b⁴
```
Subtract the third product from the third dividend. The difference equals zero, therefore we've completed the process. If the first term of the difference, Dividend4, does not contain the first term of the divisor, we've also completed the process.
Dividend3 - P3 = 0 (Dividend4)
```
a²b² + 2ab³ +  4b⁴ minus
a²b² + 2ab³ + 4b⁴
_________________
0
```

Page 58-60

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Divide Polynomials

Algebra Help: Section 2.22

Rules

Simplified Steps

Help with Algebra Homework

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