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In addition of algebraic expressions while adding algebraic expressions we collect the like terms and add them. The sum of several like terms is the like term whose coefficient is the sum of the coefficients of these like terms.

Two ways to solve addition of algebraic expressions.

Horizontal Method: In this method, all expressions are written in a horizontal line and then the terms are arranged to collect all the groups of like terms and then added.

Column Method: In this method each expression is written in a separate row such that there like terms are arranged one below the other in a column. Then the addition of terms is done column wise.

Following illustrations will illustrate these methods.

Examples on addition of algebraic expressions:

3. Add: 8x² - 5xy + 3y², 2xy - 6y² + 3x² and y² + xy - 6x².

Solution:

Arranging the given expressions in descending powers of x with like terms under each other and adding column wise;

8x² - 5xy + 3y²

3x² - 2xy - 6y²

-6x² + xy + y²
_____________
5x² - 2xy - 2y²
_____________

= 5x² - 2xy - 2y²

4. Add: 11a² + 8b² - 9c², 5b² + 3c² - 4a² and 3a² - 4b² - 4c².

Solution:

Writing the terms of the given expressions in the same order in form of rows with like terms below each other and adding column wise;

11a² + 8b² - 9c²

- 4a² + 5b² + 3c²

3a² - 4b² - 4c²
________________
10a² + 9b² - 10c²
________________

= 10a² + 9b² - 10c²

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Addition of Algebraic Expressions Subtraction of Algebraic Expressions Multiplication of Algebraic Expression Division of Algebraic Expressions 8th Grade Math Practice

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Addition of Algebraic Expressions |Collect the Like Terms|Algebraic Expressions

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In addition of algebraic expressions while adding algebraic expressions we collect the like terms and add them. The sum of several like terms is the like term whose coefficient is the sum of the coefficients of these like terms. Two ways to solve addition of algebraic expressions.

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1.Algebra Tutor
2.The Fun Way to Learn Algebra

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Subtraction of algebraic expressions are explained in each steps:

Steps I: Arrange the terms of the given expressions in the same order.

Steps II: Write the given expressions in two rows in such a way that the like terms occur one below the other, keeping the expression to be subtracted in the second row.

Steps III: Change the sign of each term in the lower row from + to - and from - to +.

Steps IV: With new signs of the terms of lower row, add column wise.

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Addition of Algebraic Expressions Subtraction of Algebraic Expressions Multiplication of Algebraic Expression Division of Algebraic Expressions 8th Grade Math Practice

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Subtraction of Algebraic Expressions | Subtracting Column-wise | Math Only Math

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Subtraction of algebraic expressions are explained in each steps: Steps I: Arrange the terms of the given expressions in the same order. Steps II: Write the given expressions in two rows in such a way that the like terms occur one below the other, keeping the expression to be subtracted in the second row.

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Multiplication of Algebraic Expression | Product of Algebraic Expressions

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In multiplication of algebraic expression before taking up the product of algebraic expressions, let us look at two simple rules. (i) The product of two factors with like signs is positive, and

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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:
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:

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).

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

2. Divide 2x² + 3x + 1 by (x + 1).

3. Divide x² + 6x + 8 by (x + 4).

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

Solution:

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

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

Solution:

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

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,

7. Using division, show that (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,

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;

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Addition of Algebraic Expressions Subtraction of Algebraic Expressions Multiplication of Algebraic Expression Division of Algebraic Expressions 8th Grade Math Practice

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Division of Algebraic Expression | Division of a Monomial, Polynomial

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In division of algebraic expression if x is a variable and m, n are positive integers such that m > n then (x m ÷ x n ) = x m - n . I. Division of a Monomial by a Monomial

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Online quiz

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Operations with Algebraic Expressions mathfm

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Operations with Algebraic Expressions mathfm