# Division of Algebraic Expression

In division of algebraic expression if x is a variable and m, n are positive integers such that m > n then (xᵐ ÷ xⁿ) = x$$^{m - n}$$.

### I. Division of a Monomial by a Monomial

Quotient of two monomials is a monomial which is equal to the quotient of their numerical coefficients, multiplied by the quotient of their literal coefficients.

Rule:

Quotient of two monomials = (quotient of their numerical coefficients) x (quotient of their variables)

### Divide:

(i) 8x2y3 by -2xy

Solution:

(i) 8x2y3/-2xy

= (8/-2) x2 - 1y3 - 1 [Using quotient law xm ÷ xn = xm - n]

= -4xy2.

(ii) 35x3yz2 by -7xyz

Solution:

35x3yz2 by -7xyz

= (35/-7) x3 - 1y1 - 1z2 - 1 [Using quotient law xm ÷ xn = xm - n]

= -5 x2y0z1 [y0 = 1]

= -5x2z.

(iii) -15x3yz3 by -5xyz2

Solution:

-15x3yz3 by -5xyz2.

= (-15/-5) x3 - 1y1 - 1z3 - 2. [Using quotient law xm ÷ xn = xm - n].

= 3 x2y0z1 [y0 = 1].

= 3x2z.

### II. Division of a Polynomial by a Monomial

Rule:

For dividing a polynomial by a monomial, divide each term of the polynomial by the monomial. We divide each term of the polynomial by the monomial and then simplify.

### Divide:

(i) 6x5 + 18x4 - 3x2 by 3x2

Solution:

6x5 + 18x4 - 3x2 by 3x2

= (6x5 + 18x4 - 3x2) ÷ 3x2 6x5/3x2 + 18x4/3x2 - 3x2/3x2

=2x3 + 6x2 - 1.

(ii) 20x3y + 12x2y2 - 10xy by 2xy

Solution:

20x3y + 12x2y2 - 10xy by 2xy

= (20x3y + 12x2y2 - 10xy) ÷ 2xy

= 20x3y/2xy + 12x2y2/2xy - 10xy/2xy

= 10x2 + 6xy - 5.

### III. Division of a Polynomial by a Polynomial

We may proceed according to the steps given below:

(i) Arrange the terms of the dividend and divisor in descending order of their degrees.

(ii) Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.

(iii) Multiply all the terms of the divisor by the first term of the quotient and subtract the result from the dividend.

(iv) Consider the remainder (if any) as a new dividend and proceed as before.

(v) Repeat this process till we obtain a remainder which is either 0 or a polynomial of degree less than that of the divisor.

Let us understand it through some examples.

### 1. Divide 12 – 14a² – 13a by (3 + 2a).

Solution:

12 – 14a² – 13a by (3 + 2a).

Write the terms of the polynomial (dividend and divisor both) in decreasing order of exponents of variables.

So, dividend becomes – 14a² – 13a + 12 and divisor becomes 2a + 3.

Divide the first term of the dividend by the first term of the divisor which gives first term of the quotient.

Multiply the divisor by the first term of the quotient and subtract the product from the dividend which gives the remainder.

Now, this remainder is treated as, new dividend but the divisor remains the same.

Now, we divide the first term of the new dividend by the first term of the divisor which gives second term of the quotient.

Now, multiply the divisor by the term of the quotient just obtained and subtracts the product from the dividend.

Thus, we conclude that divisor and quotient are the factors of dividend if the remainder is zero.

Quotient = -7a + 4

Remainder = 0

### Dividend = divisor × quotient + remainder

= (2a + 3)(-7a + 4) + 0

= 2a(-7a + 4) +3(-7a + 4) + 0

= – 14a² + 8a – 21a + 12 + 0

= – 14a² – 13a + 12

### 2. Divide 2x² + 3x + 1 by (x + 1).

Solution:

Therefore, quotient = (2x + 1) and remainder = 0.

### 3. Divide x² + 6x + 8 by (x + 4).

Solution:

Therefore, Dividend = x² + 6x + 8

Divisor = x + 4

Quotient = x + 2 and

Remainder = 0.

### 4. Divide 9x - 6x² + x³ - 2 by (x - 2).

Solution:

Arranging the terms of the dividend and divisor in descending order and then dividing,

Therefore, quotient = (x² - 4x + 1) and remainder = 0.

### 5. Divide (29x - 6x² - 28) by (3x -4).

Solution:

Arranging the terms of the dividend and divisor in descending order and then dividing,

Therefore, (29x - 6x² - 28) ÷ (3x - 4) = (-2x + 7).

### 6. Divide (5x³ - 4x² + 3x + 18) by (3 - 2x + x²).

Solution:

The terms of the dividend are in descending order.

Arranging the terms of the divisor in descending order and then dividing,

Therefore, (5x³ - 4x² + 3x + 18) ÷ (x² - 2x + 3) = (5x + 6).

### 7. Using division, show that (x - 1) is a factor of (x³ - 1).

Solution:

(x - 1) completely divides (x³ - 1).

Hence, (x - 1) is a factor of(x³- 1).

### 8. Find the quotient and remainder when (7 + 15x - 13x² + 5x³) is divided by (4 - 3x + x²).

Solution:

Arranging the terms of dividend and divisor in descending order and then dividing,

Therefore, quotient is (5x + 2) and remainder is (x - 1).

### 9. Divide (10x⁴ + 17x³ - 62x² + 30x - 3) by (2x² + 7x - 1).

Solution:

The terms of the dividend and that of the divisor are in descending order. So, we divide them as;

(10x⁴ + 17x³ - 62x² + 30x - 3) ÷ (2x² + 7x - 1) = (5x² - 9x + 3).

Algebraic Expression

Algebraic Expression

Division of Algebraic Expressions

### New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

 Share this page: What’s this?

## Recent Articles

1. ### Roman Numerals | System of Numbers | Symbol of Roman Numerals |Numbers

Feb 22, 24 04:21 PM

How to read and write roman numerals? Hundreds of year ago, the Romans had a system of numbers which had only seven symbols. Each symbol had a different value and there was no symbol for 0. The symbol…

Read More

2. ### Worksheet on Roman Numerals |Roman Numerals|Symbols for Roman Numerals

Feb 22, 24 04:15 PM

Practice the worksheet on roman numerals or numbers. This sheet will encourage the students to practice about the symbols for roman numerals and their values. Write the number for the following: (a) V…

Read More

3. ### Roman Symbols | What are Roman Numbers? | Roman Numeration System

Feb 22, 24 02:30 PM

Do we know from where Roman symbols came? In Rome, people wanted to use their own symbols to express various numbers. These symbols, used by Romans, are known as Roman symbols, Romans used only seven…

Read More

4. ### Place Value | Place, Place Value and Face Value | Grouping the Digits

Feb 19, 24 11:57 PM

The place value of a digit in a number is the value it holds to be at the place in the number. We know about the place value and face value of a digit and we will learn about it in details. We know th…

Read More

5. ### Math Questions Answers | Solved Math Questions and Answers | Free Math

Feb 19, 24 11:14 PM

In math questions answers each questions are solved with explanation. The questions are based from different topics. Care has been taken to solve the questions in such a way that students

Read More