# 8.10: Dividing Polynomials

### Dividing A Polynomial By A Monomial

The following examples illustrate how to divide a polynomial by a monomial. The division process is quite simple and is based on the addition of rational expressions.

(dfrac{a}{c} + dfrac{b}{c} = dfrac{a+b}{c})

Turning this equation around we get

(dfrac{a+b}{c} = dfrac{a}{c} + dfrac{b}{c})

Now we simply divide (c) into (a), and (c) into (b). This should suggest a rule.

Dividing a Polynomial By a Monomial

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

### Sample Set A

Example (PageIndex{1})

(dfrac{3x^2 + x - 11}{x}). Divide every term of (3x^2 + x - 11) by (x).

(dfrac{3x^2}{x} + dfrac{x}{x} - dfrac{11}{x} = 3x + 1 - dfrac{11}{x})

Example (PageIndex{2})

(dfrac{8x^3 + 4a^2 - 16a + 9}{2a^2}. Divide every term of (8a^3 + 4a^2 - 16a + 9) by (2a^2).

Example (PageIndex{3})

(dfrac{4b^6 - 9b^4 - 2b + 5}{-4b^2}). Divide every term of (4b^6 - 9b^4 - 2b + 5) by (-4b^2).

(dfrac{4b^6}{-4b^2} - dfrac{9b^4}{-4b^2} - dfrac{2b}{-4b^2} + dfrac{5}{-4b^2} = -b^4 + dfrac{9}{4}b^2 + dfrac{1}{2b} - dfrac{5}{4b^2})

### Practice Set A

Perform the following divisions.

Practice Problem (PageIndex{1})

(dfrac{2x^2 + x - 1}{x})

(2x + 1 - dfrac{1}{x})

Practice Problem (PageIndex{2})

(dfrac{3x^3 + 4x^2 + 10x - 4}{x^2})

(3x + 4 + dfrac{10}{x} - dfrac{4}{x^2})

Practice Problem (PageIndex{3})

(dfrac{a^2b + 3ab^2 + 2b}{ab})

(a + 3b + dfrac{2}{a})

Practice Problem (PageIndex{4})

(dfrac{14x^2y^2 - 7xy}{7xy})

(2xy−1)

Practice Problem (PageIndex{5})

(dfrac{10m^3n^2 + 15m^2n^3 - 20mn}{-5m})

(-2m^2n^2 - 3mn^3 + 4n)

### The Process Of Division

In Section 8.3 we studied the method of reducing rational expressions. For example, we observed how to reduce an expression such as

(dfrac{x^2 - 2x - 8}{x^2 - 3x - 4})

Our method was to factor both the numerator and denominator, then divide out common factors.

(dfrac{(x-4)(x+2)}{(x-4)(x+1)})

(dfrac{cancel{(x-4)}(x+2)}{cancel{(x-4)}(x+1)})

(dfrac{x+2}{x+1})

When the numerator and denominator have no factors in common, the division may still occur, but the process is a little more involved than merely factoring. The method of dividing one polynomial by another is much the same as that of dividing one number by another. First, we’ll review the steps in dividing numbers.

(dfrac{35}{8}). We are to divide 35 by 8.

We try 4, since 32 divided by 8 is 4.

Multiply 4 and 8

Subtract 32 from 35

Since the remainder 3 is less than the divisor 8, we are done with the 32 division.

(4dfrac{3}{8}). The quotient is expressed as a mixed number.

The process was to divide, multiply, and subtract.

### Review Of Subtraction Of Polynomials

A very important step in the process of dividing one polynomial by another is the subtraction of polynomials. Let’s review the process of subtraction by observing a few examples.

1. Subtract (x -2) from (x-5); that is, find ((x-5) - (x-2)).

Since (x-2) is preceded by a minus sign, remove the parentheses, change the sign of each term, then add.

(egin{array}{flushleft}
x-5 && x-5
-(x-2) && -x+2
ext{_______} & = & ext{_______}
&&-3
end{array})

The result is (-3)

2. Subtract (x^3 + 3x^2) from (x^3 + 4x^2 + x - 1).

Since (x^3 + 3x^2) is preceded by a minus sign, remove the parentheses, change the sign of each term, then add.

(egin{array}{flushleft}
x^3 + 4x^2 + x - 1 && x^3 + 4x^2 + x - 1
-(x^3 + 3x^2) && -x^3 - 3x^2
ext{_______________} & = & ext{_______________}
&&x^2 + x - 1
end{array})

The result is (x^2 + x - 1)

3. Subtract (x^2 + 3x) from (x^2 + 1)

We can write (x^2 + 1) as (x^2 + 0x + 1).

(egin{array}{flushleft}
x^2 + 1 && x^2 + 0x + 1 && x^2 + 0x + 1
-(x^2 + 3x) && -(x^2 + 3x) && -x^2 - 3x
ext{____________} & = & ext{____________} & = & ext{____________}
&&&& -3x + 1
end{array})

### Dividing A Polynomial By A Polynomial

Now we’ll observe some examples of dividing one polynomial by another. The process is the same as the process used with whole numbers: divide, multiply, subtract, divide, multiply, subtract,....

The division, multiplication, and subtraction take place one term at a time. The process is concluded when the polynomial remainder is of lesser degree than the polynomial divisor.

### Sample Set B

Perform the division.

Example (PageIndex{4})

(dfrac{x-5}{x-2}). We are to divide (x-5) by (x-2).

(1 - dfrac{3}{x-2})

Thus,

(dfrac{x-5}{x-2} = 1 - dfrac{3}{x-2})

Example (PageIndex{5})

(dfrac{x^3 + 4x^2 + x - 1}{x + 3}). We are to divide (x^3 + 4x^2 + x - 1) by (x + 3).

(x^2 + x - 2 + dfrac{5}{x+3})

Thus,

(dfrac{x^3 + 4x^2 + x - 1}{x + 3} = x^2 + x - 2 + dfrac{5}{x+3})

### Practice Set B

Perform the following divisions.

Practice Problem (PageIndex{6})

(dfrac{x+6}{x-1})

(1 + dfrac{7}{x-1})

Practice Problem (PageIndex{7})

(dfrac{x^2 + 2x + 5}{x + 3})

(x - 1 + dfrac{8}{x+3})

Practice Problem (PageIndex{8})

(dfrac{x^3 + x^2 - x - 2}{x + 8})

(x^2 - 7x + 55 - dfrac{442}{x+8})

Practice Problem (PageIndex{9})

(dfrac{x^3 + x^2 - 3x + 1}{x^2 + 4x - 5})

(x - 3 + dfrac{14x - 14}{x^2 + 4x - 5} = x - 3 + dfrac{14}{x+5})

### Sample Set C

Example (PageIndex{6})

Divide (2x^3 - 4x + 1) by (x + 6)

(dfrac{2x^3 - 4x + 1}{x + 6}) Notice that the (x^2) term in the numerator is missing. We can avoid any confusion by writing

(dfrac{2x^3 + 0x^2 - 4x + 1}{x+6}) Divide, multiply, and subtract.

(dfrac{2x^3 - 4x + 1}{x + 6} = 2x^3 - 12x + 68 - dfrac{407}{x + 6})

### Practice Set C

Perform the following divisions.

Practice Problem (PageIndex{10})

(dfrac{x^2 - 3}{x+2})

(x - 2 + dfrac{1}{x+2})

Practice Problem (PageIndex{11})

(dfrac{4x^2 - 1}{x-3})

(4x + 12 + dfrac{35}{x-3})

Practice Problem (PageIndex{12})

(dfrac{x^3 + 2x + 2}{x-2})

(x^2 + 2x + 6 + dfrac{14}{x-2})

Practice Problem (PageIndex{13})

(dfrac{6x^3 + 5x^2 - 1}{2x + 3})

(3x^2 - 2x + 3 - dfrac{10}{2x + 3})

### Exercises

For the following problems, perform the divisions.

Exercise (PageIndex{1})

(dfrac{6a + 12}{2})

(3a+6)

Exercise (PageIndex{2})

(dfrac{12b - 6}{3})

Exercise (PageIndex{3})

(dfrac{8y - 4}{-4})

(−2y+1)

Exercise (PageIndex{4})

(dfrac{21a - 9}{-3})

Exercise (PageIndex{5})

(dfrac{3x^2 - 6x}{-3})

(−x(x−2))

Exercise (PageIndex{6})

(dfrac{4y^2 - 2y}{2y})

Exercise (PageIndex{7})

(dfrac{9a^2 + 3a}{2a})

(3a+1)

Exercise (PageIndex{8})

(dfrac{20x^2 + 10x}{5x})

Exercise (PageIndex{9})

(dfrac{6x^3 + 2x^2 + 8x}{2x})

(3x^2 + x + 4)

Exercise (PageIndex{10})

(dfrac{26y^3 + 13y^2 + 39y}{13y})

Exercise (PageIndex{11})

(dfrac{a^2b^2 + 4a^2b + 6ab^2 - 10ab}{ab})

(ab+4a+6b−10)

Exercise (PageIndex{12})

(dfrac{7x^3y + 8x^2y^3 + 3xy^4 - 4xy}{xy})

Exercise (PageIndex{13})

(dfrac{5x^3y^3 - 15x^2y^2 + 20xy}{-5xy})

(-x^2y^2 + 3xy - 4)

Exercise (PageIndex{14})

(dfrac{4a^2b^3 - 8ab^4 + 12ab^2}{-2ab^2})

Exercise (PageIndex{15})

(dfrac{6a^2y^2 + 12a^2y + 18a^2}{24a^2})

(dfrac{1}{4}y^2 + dfrac{1}{2}y + dfrac{3}{4})

Exercise (PageIndex{16})

(dfrac{3c^3y^3 + 99c^3y^4 - 12c^3y^5}{3x^3y^3})

Exercise (PageIndex{17})

(dfrac{16ax^2 - 20ax^3 + 24ax^4}{6a^4})

(dfrac{8x^2 - 10x^3 + 12x^4}{3a^3}) or (dfrac{12x^4 - 10x^3 + 8x^2}{3a^2})

Exercise (PageIndex{18})

(dfrac{21ay^3 - 18ay^2 - 15ay}{6ay^2})

Exercise (PageIndex{19})

(dfrac{-14b^2c^2 + 21b^3 - 28c^3}{-7a^2c^3})

(dfrac{2b^2 - 3b^3c + 4c}{a^2c})

Exercise (PageIndex{20})

(dfrac{-30a^2b^4 - 35a^2b^3 - 25a^2}{-5b^3})

Exercise (PageIndex{21})

(dfrac{x+6}{x-2})

(1 + dfrac{8}{x-2})

Exercise (PageIndex{22})

(dfrac{y + 7}{y + 1})

Exercise (PageIndex{23})

(dfrac{x^2 - x + 4}{x + 2})

(x - 3 + dfrac{10}{x+2})

Exercise (PageIndex{24})

(dfrac{x^2 + 2x - 1}{x + 1})

Exercise (PageIndex{25})

(dfrac{x^2 - x + 3}{x + 1})

(x - 2 + dfrac{5}{x + 1})

Exercise (PageIndex{26})

(dfrac{x^2 + 5x + 5}{x + 5})

Exercise (PageIndex{27})

(dfrac{x^2 - 2}{x + 1})

(x - 1 - dfrac{1}{x+1})

Exercise (PageIndex{28})

(dfrac{a^2 - 6}{a + 2})

Exercise (PageIndex{29})

(dfrac{y^2 + 4}{y + 2})

(y - 2 + dfrac{8}{y + 2})

Exercise (PageIndex{30})

(dfrac{x^2 + 36}{x + 6})

Exercise (PageIndex{31})

(dfrac{x^3 - 1}{x + 1})

(x^2 - x + 1 - dfrac{2}{x + 1})

Exercise (PageIndex{32})

(dfrac{a^3 - 8}{a + 2})

Exercise (PageIndex{33})

(dfrac{x^3 + 3x^2 + x - 2}{x-2})

(x^2 + 5x + 11 + dfrac{20}{x-2})

Exercise (PageIndex{34})

(dfrac{a^3 + 2a^2 - a + 1}{a - 3})

Exercise (PageIndex{35})

(dfrac{x^3 + 2x + 1}{x - 3})

Exercise (PageIndex{36})

(dfrac{y^3 + 2y^2 + 4}{y + 2})

(y^2 + y - 2 + dfrac{8}{y + 2})

Exercise (PageIndex{37})

(dfrac{y^3 + 5y^2 - 3}{y - 1})

Exercise (PageIndex{38})

(dfrac{x^3 + 3x^2}{x + 3})

(x^2)

Exercise (PageIndex{39})

(dfrac{a^2 + 2a}{a + 2})

Exercise (PageIndex{40})

(dfrac{x^2 - x - 6}{x^2 - 2x - 3})

(1 + dfrac{1}{x + 1})

Exercise (PageIndex{41})

(dfrac{a^2 + 5a + 4}{a^2 - a - 2})

Exercise (PageIndex{42})

(dfrac{2y^2 + 5y + 3}{y^2 - 3y - 4})

(2 + dfrac{11}{y-4})

Exercise (PageIndex{43})

(dfrac{3a^2 + 4a + 2}{3a + 4})

Exercise (PageIndex{44})

(dfrac{6x^2 + 8x - 1}{3x + 4})

(2x - dfrac{1}{3x + 4})

Exercise (PageIndex{45})

(dfrac{20y^2 + 15y - 4}{4y + 3})

Exercise (PageIndex{46})

(dfrac{4x^3 + 4x^2 - 3x - 2}{2x - 1})

(2x^2 + 3x - dfrac{2}{2x - 1})

Exercise (PageIndex{47})

(dfrac{9a^3 - 18a^2 8a - 1}{3a - 2})

Exercise (PageIndex{48})

(dfrac{4x^4 - 4x^3 + 2x^2 - 2x - 1}{x-1})

(4x^3 + 2x - dfrac{1}{x-1})

Exercise (PageIndex{49})

(dfrac{3y^4 + 9y^3 - 2y^2 - 6y + 4}{y + 3})

Exercise (PageIndex{50})

(dfrac{3y^2 + 3y + 5}{y^2 + y + 1})

(3 + dfrac{2}{y^2 + y + 1})

Exercise (PageIndex{51})

(dfrac{2a^2 + 4a + 1}{a^2 + 2a + 3})

Exercise (PageIndex{52})

(dfrac{8z^6 - 4z^5 - 8z^4 + 8z^3 + 3z^2 - 14z}{2z - 3})

(4z^5 + 4z^4 + 2z^3 + 7z^2 + 12z + 11 + dfrac{33}{2z - 3})

Exercise (PageIndex{53})

(dfrac{9 a^{7}+15 a^{6}+4 a^{5}-3 a^{4}-a^{3}+12 a^{2}+a-5}{3 a+1})

Exercise (PageIndex{54})

((2x^5 + 5x^4 -1) div (2x + 5))

(x^4 - dfrac{1}{2x + 5})

Exercise (PageIndex{55})

((6a^4 - 2a^3 - 3a^2 + a + 4) div (3a - 1))

### Exercises For Review

Exercise (PageIndex{56})

Find the product. (dfrac{x^2 + 2x - 8}{x^2 - 9} cdot dfrac{2x + 6}{4x - 8})

(dfrac{x + 4}{2(x-3)})

Exercise (PageIndex{57})

Find the sum. (dfrac{x-7}{x + 5} + dfrac{x + 4}{x - 2})

Exercise (PageIndex{58})

Solve the equation (dfrac{1}{x + 3} + dfrac{1}{x - 3} = dfrac{1}{x^2 - 9})

(x = dfrac{1}{2})

Exercise (PageIndex{59})

When the same number is subtracted from both the numerator and denominator of (dfrac{3}{10}), the result is (dfrac{1}{8}). What is teh number that is subtracted?

Exercise (PageIndex{60})

Simplify (dfrac{frac{1}{x+5}}{frac{4}{x^{2}-25}})

(dfrac{x-5}{4})

## Maharashtra Board Class 8 Maths Solutions Chapter 10 Division of Polynomials Practice Set 10.2

Division of Polynomials Class 8 Practice Set 10.2 Question 1. Divide and write the quotient and the remainder.
i. (y 2 + 10y + 24) ÷ (y + 4)
ii. (p 2 + 7p – 5) ÷ (p + 3)
iii. (3x + 2x 2 + 4x 3 ) ÷ (x – 4)
iv. (2m 3 + m 2 + m + 9) ÷ (2m – 1)
v. (3x – 3x 2 – 12 + x 4 + x 3 ) ÷ (2 + x 2 )
vi. (a 4 – a 3 + a 2 – a + 1) ÷ (a 3 – 2)
vii. (4x 4 – 5x 3 – 7x + 1) ÷ (4x – 1)
Solution:
i. (y 2 + 10y + 24) ÷ (y + 4)

∴ Quotient = y + 6
Remainder = 0

ii. (p 2 + 7p – 5) ÷ (p + 3)

∴ Quotient = p + 4
Remainder = -17

iii. (3x + 2x 2 + 4x 3 ) ÷ (x – 4)
Write the dividend in descending order of their indices.
3x + 2x² + 4x³ = 4x³ + 2x² + 3x

∴ Quotient = 4x² + 18x + 75
Remainder = 300

iv. (2m 3 + m 2 + m + 9) ÷ (2m – 1)

∴ Quotient = m² + m + 1
Remainder = 10

v. (3x – 3x 2 – 12 + x 4 + x 3 ) ÷ (2 + x 2 )
Write the dividend in descending order of their indices.
(x 4 + x 3 – 3x 2 + 3x – 12) ÷ (x 2 + 2)

∴ Quotient = x² + x – 5
Remainder = x – 2

vi. (a 4 – a 3 + a 2 – a + 1) ÷ (a 3 – 2)

∴ Quotient = a – 1
Remainder = a² + a – 1

vii. (4x 4 – 5x 3 – 7x + 1) ÷ (4x – 1)
Write the dividend in descending order of their indices.
(4x 4 – 5x 3 – 7x + 1) = (4x 4 – 5x 3 + 0x 2 – 7x + 1)

∴ Quotient = (x^<3>-x^<2>-frac<4>-frac<29><16>)
Remainder = (frac < -13 >< 16 >)

## Dividing by a Polynomial

The same technique outlined for dividing by a monomial does not work for polynomials with two or more terms in the denominator. In this section, we will outline a process called polynomial long division The process of dividing two polynomials using the division algorithm. , which is based on the division algorithm for real numbers. For the sake of clarity, we will assume that all expressions in the denominator are nonzero.

Example 5: Divide: x 3 + 3 x 2 − 8 x − 4 x − 2 .

Solution: Here x − 2 is the divisor and x 3 + 3 x 2 − 8 x − 4 is the dividend.

Step 1: To determine the first term of the quotient, divide the leading term of the dividend by the leading term of the divisor.

Step 2: Multiply the first term of the quotient by the divisor, remembering to distribute, and line up like terms with the dividend.

Step 3: Subtract the resulting quantity from the dividend. Take care to subtract both terms.

Step 4: Bring down the remaining terms and repeat the process from step 1.

Notice that the leading term is eliminated and that the result has a degree that is one less than the dividend. The complete process is illustrated below:

Polynomial long division ends when the degree of the remainder The expression that is left after the division algorithm ends. is less than the degree of the divisor. Here the remainder is 0. Therefore, the binomial divides the polynomial evenly and the answer is the quotient shown above the division line.

To check the answer, multiply the divisor by the quotient to see if you obtain the dividend:

Next, we demonstrate the case where there is a nonzero remainder.

Just as with real numbers, the final answer adds the fraction where the remainder is the numerator and the divisor is the denominator to the quotient. In general, when dividing we have

If we multiply both sides by the divisor we obtain

Example 6: Divide: 6 x 2 − 5 x + 3 2 x − 1 .

Solution: Since the denominator is a binomial, begin by setting up polynomial long division.

To start, determine what monomial times 2 x − 1 results in a leading term 6 x 2 . This is the quotient of the given leading terms: ( 6 x 2 ) ÷ ( 2 x ) = 3 x . Multiply 3 x times the divisor 2 x − 1 and line up the result with like terms of the dividend.

Subtract the result from the dividend and bring down the constant term +3.

Subtracting eliminates the leading term and − 5 x − ( − 3 x ) = − 5 x + 3 x = − 2 x . The quotient of − 2 x and 2 x is −1. Multiply 2 x − 1 by −1 and line up the result.

Subtract again and notice that we are left with a remainder.

The constant term 2 has degree 0, and thus the division ends. We may write

Answer: 3 x − 1 + 2 2 x − 1 . To check that this result is correct, we multiply as follows:

Occasionally, some of the powers of the variables appear to be missing within a polynomial. This can lead to errors when lining up like terms. Therefore, when first learning how to divide polynomials using long division, fill in the missing terms with zero coefficients, called placeholders Terms with zero coefficients used to fill in all missing exponents within a polynomial. .

Example 7: Divide: 27 x 3 + 64 3 x + 4 .

Solution: Notice that the binomial in the numerator does not have terms with degree 2 or 1. The division is simplified if we rewrite the expression with placeholders:

Set up polynomial long division:

We begin with 27 x 3 ÷ 3 x = 9 x 2 and work the rest of the division algorithm.

Example 8: Divide: 3 x 4 − 2 x 3 + 6 x 2 + 23 x − 7 x 2 − 2 x + 5 .

Begin the process by dividing the leading terms to determine the leading term of the quotient 3 x 4 ÷ x 2 = 3 x 2 . Take care to distribute and line up the like terms. Continue the process until the remainder has a degree less than 2.

The remainder is x − 2 . Write the answer with the remainder:

Answer: 3 x 2 + 4 x − 1 + x − 2 x 2 − 2 x + 5

Polynomial long division takes time and practice to master. Work lots of problems and remember that you may check your answers by multiplying the quotient by the divisor (and adding the remainder if present) to obtain the dividend.

Try this! Divide: 20 x 4 − 32 x 3 + 7 x 2 + 8 x − 10 5 x − 3 .

Answer: 4 x 3 − 4 x 2 − x + 1 − 7 5 x − 3

### Video Solution

A main topic in algebra classes is polynomials. There are many subtopics of this topic, including adding, subtracting, multiplying, and dividing.

This page will take you through the basics, including the definition, labeling according to the number of terms, labeling according to the degree, and the end behavior.

Definition: A monomial or a sum of monomials. A monomial is simply a "term."

Terms are seperated by + and - signs. Poly's with 1, 2, or 3 terms have specific names, while poly's of 4 or more terms are simply called polynomials of # terms. Take a look at the table below.

Again, all you are looking for is the number of terms. When there are 1, 2, or 3 the poly is given the special names monomial, binomial, trinomial, respectively. If there are more than 3, then it is simply called a poly of that many terms.

 Labeling According to the Degree

The degree of a poly is determined by the exponents. In fact, it is determined by the largest exponent. Take a look at this table.

Notice that we are no longer interested in the number of terms, but simply the degree of the exponent.

Often polynomials will be referred to by both the number of terms and the degree. Take a look at the two examples below.

 Example #1: x 2 + 4x - 8 (quadratic trinomial) Example #2: 2x 3 - 16x (cubic binomial)

 Standard Form

For simplicity, it is often preferred to put a poly into standard form. This means that the terms are place in descending order (highest to lowest) according to their degree.

 End Behavior of the Polynomial

The end behavior of a graph is what happens at the far left and the far right. Two factors determine the end behavior: positive or negative, and whether the degree is even or odd.

For the examples below, we will use x 2 and x 3 , but the end behavior will be the same for any even degree or any odd degree. However, keep in mind that what happens "in the middle" will change.

Knowing the general shape and end behavior is an important step towards understanding polys. With a little practice, you will know how to perform all of the operations associated with polynomials. Take a look at how to perform polynomials division.

## What Are the Steps for Synthetic Division of Polynomials?

The following are the steps while performing synthetic division and finding the quotient and the remainder.

• Check whether the polynomial is in the standard form.
• Write the coefficients in the dividend's place.
• Write the zero of the linear factor in the divisor's place.
• Bring the first coefficient down.
• Multiply it with the divisor and write it below the next coefficient.
• Add them and write the value below.
• Repeat the previous 2 steps until you reach the last term.
• Separate the last term thus obtained which is the remainder.
• Now group the coefficients with the variables to get the quotient.

### Synthetic Division Calculator

In the following simulation, let's recall the technique to divide using synthetic division.

Choose the order of the polynomial and enter the coefficients of the dividend polynomial and give the value of the linear factor.

You can also verify your resultant polynomial using the long division method.

• Write down the coefficients and divide them using the zero of the linear factor to obtain the quotient and the remainder. (dfrac<(x-a)>= Q(x) + dfrac)
• When we do synthetic division by (bx + a), we should get (dfrac) as the quotient.

Outline: If the constant term of the polynomial is $, the result is obvious. The rest of the proof imitates the standard Euclid-style proof that there are infinitely many primes. So let the constant term be$a e 0$. It follows that the polynomial$g(n)$has the shape$g(n)=nq(n)+a,$where$q(n)$is a polynomial with integer coefficients. As$n$gets large,$g(n)$becomes very large positive or very large negative. Without loss of generality we can assume it becomes very large positive. In particular, for$n$large enough we have$g(n)gt |a|$. Now let$k$be large, and look at$g(k!a^2)=a^2k!q(a^2k!)+a=a(ak!q(a^2k!)+1)$. Then$ak!q(a^2k!)+1$is divisible by some prime, and that prime must be greater than$k\$.

Last reply by: Professor Eric Smith
Mon May 4, 2020 5:30 PM

Post by Sylvia Wang on April 24, 2020

how do we know where to put a place holder and how many

Last reply by: Professor Eric Smith
Wed Apr 1, 2020 12:28 PM

Post by Orlando Cao on March 31, 2020

for example 3, the answer can simplify to x + 4 + 1/2x+3.

Last reply by: Professor Eric Smith
Wed Apr 1, 2020 12:26 PM

Post by Sha Tao on March 31, 2020

Didn't you say polynomials couldn't have the variable as denominator?

Last reply by: Professor Eric Smith
Sat Mar 2, 2019 4:02 PM

Post by Kenneth Geller on November 29, 2018

The query I asked was more to see if there was anyone "alive" yet, as it seems questions go back to 2013-2014. I just finished all of Algebra 1 and I think you are an excellent teacher. I have always enjoyed math when in H.S. some 55 years ago and now that I am retired have to do something to keep the brain stimulated. One would think mathematics would not change in that time, but I do not recall division of polynomials, nor imaginary numbers. Anyway your course has been a pleasure, your students are fortunate.

Last reply by: Professor Eric Smith
Sat Nov 17, 2018 12:48 PM

Post by Kenneth Geller on October 17, 2018

Last reply by: Professor Eric Smith
Mon Jul 31, 2017 10:29 AM

Post by Joselin ji on July 30, 2017

The pause button doesn't work, and neither does the scrubbing bar on the bottom. Does anyone know why?

Post by Khanh Nguyen on October 12, 2015

Professor, there are many problems on the practice questions regarding the help given in the "show next step" button.

Could you, or someone authorized fix it? :^D

Post by patrick guerin on July 11, 2014

Last reply by: Professor Eric Smith
Sun Jul 6, 2014 2:31 PM

Post by David Saver on July 3, 2014

You really make things easy to understand!
Thanks!!

Last reply by: Professor Eric Smith
Tue Aug 20, 2013 2:20 PM

Post by Rana Laghaei on August 19, 2013

Thanks professor I really like your teaching style.You explain everything well and why it works.I hope you do an algebra 2 course.I appreciate your work!

Post by Professor Eric Smith on August 12, 2013

You should put in a zero place holder any time you have a missing power in the polynomial. For example, if you are dividing by x^3 + 2x - 1, then you want to put in a 0x^2 for the missing x squared term. This is the process you are seeing here in example 4. In this example the x squared term, and the single x term are both missing so we put in a place holder for each of them. Let me know if that helps out.

## DIVIDING POLYNOMIALS USING LONG DIVISION

where r (x) = 0 or degree of r (x) < degree of g(x) .

The polynomial p(x) is the dividend, g(x) is the divisor, q(x) is the quotient and r (x)  is the remainder.

(1) ==> Dividend = (Divisor x Quotient) + Remainder

Divide the polynomial 2x 3  - 6 x 2  + 5x + 4 by (x - 2)

Let P(x) =  2 x 3  - 6 x 2  + 5x + 4 and g(x) = x - 2

To divide the given polynomial by x - 2, we have divide the first term of the polynomial P(x) by the first term of the polynomial g(x).

If we divide  2 x 3 ਋y x, we get 2 x 2 . Now we have to multiply this  2 x 2 ਋y x - 2. From this we get  2 x 3  - 4 x 2

Now we have to subtract  2 x 3  - 4 x 2 ਏrom the given polynomial. So we get -2 x 2  + 5x + 4.

Now we have to subtract  2 x 3  - 4 x 2 ਏrom the given polynomial. So we get -2x 2  + 5x + 4.

repeat this process until we get the degree of p(x)  ≥ ꃞgree  of g(x).

Find the quotient and remainder when 4 x 3  - 5 x 2  + 6x - 2 by  x - 1.

Find the quotient and remainder when x 3  - 7x 2  - x + 6 by  x + 2.

Find the quotient and the remainder when 10- 4x + 3x 2  is divided by x - 2.

Let us first write the terms of each polynomial in descending order ( or ascending  order).

Thus, the given problem becomes  (10- 4x + 3x 2 )   ÷  (x - 2)

In the first step, we are going to divide the first term of the dividend by the first first term of the divisor.

After changing the signs, +3x 2 ਊnd -3 x 2  will get canceled. By simplifying, we get 2x + 10.

In the second step again we are going to divide the first term that is 2x by the first term of divisor that is x.

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## C++ [ edit ]

This example performs multivariate polynomial division using Buchberger's algorithm to decompose a polynomial into its Gröbner bases. Polynomials are represented as hash-maps of monomials with tuples of exponents as keys and their corresponding coefficients as values: e.g. 2xy + 3x + 5y + 7 is represented as <[1 1] 2, [1 0] 3, [0 1] 5, [0 0] 7>.

Since this algorithm is much more efficient when the input is in graded reverse lexicographic (grevlex) order a comparator is included to be used with Clojure's sorted-map— (into (sorted-map-by grevlex) . ) —as well as necessary functions to compute polynomial multiplication, monomial complements, and S-polynomials.

## Input Arguments

### P — Polynomial to divide symbolic expression | symbolic function

Polynomial to divide, specified as a symbolic expression or function.

### D — Polynomials to divide by symbolic expression or function | vector of symbolic expressions or functions

Polynomials to divide by, specified as a symbolic expression or function or a vector of symbolic expressions or functions.

### Vars — Polynomial variables vector of symbolic variables

Polynomial variables, specified as a vector of symbolic variables.

### MonomialOrder — Monomial order of divisors 'degreeInverseLexicographic' (default) | 'degreeLexicographic' | 'lexicographic'

Monomial order of divisors, specified as 'degreeInverseLexicographic' , 'degreeLexicographic' , or 'lexicographic' . If you specify vars , then polynomialReduce sorts variables based on the order of variables in vars .

lexicographic sorts the terms of a polynomial using lexicographic ordering.

degreeLexicographic sorts the terms of a polynomial according to the total degree of each term. If terms have equal total degrees, polynomialReduce sorts the terms using lexicographic ordering.

degreeInverseLexicographic sorts the terms of a polynomial according to the total degree of each term. If terms have equal total degrees, polynomialReduce sorts the terms using inverse lexicographic ordering.