Category: Mathematics

An algebraic expression is an expression that is the combination of constants and variables along with different algebraic operations such as addition, subtraction, etc. We included all types of algebraic expression problems imposed in the exams. So, students can prepare perfectly for the exam with our Algebraic Expression material. Also, find the quick links in this article where you can get the detailed concepts, questions, answers, along with explanations.

Example:

1. 2x + 3 = 6

• In 2x + 3 = 6, x is an unknown variable and
• The coefficient is assigned for the variable. 2x + 3 = 6, here 2 is the coefficient of x.
• The contant term is a definite value. 3 and 6 are the constant in 2x + 3 = 6.

Types of Algebraic Expressions

There are different types of algebraic expressions available. Let us have a look at different algebraic expressions with detailed examples.

Monomial Algebraic Expression

Monomial Algebraic Expression is an algebraic expression that contains only one term.

Example:
2x, -2xy, 3y² are some of the examples for Monomial Algebraic Expression.

Binomial Expression

Binomial Algebraic Expression is an algebraic expression that contains two terms.

Example:
6y + 8, 3y + 9, 8y³ + 2, etc. are some of the examples for Binomial Algebraic Expression.

Trinomial Expression

Trinomial Algebraic Expression is an algebraic expression that contains three terms.

Example:
2x – 3y + 6, 4x + 2y – 7z, 5a³ + 8b² + 9c⁴, etc. are some of the examples for Trinomial Algebraic Expression.

Multinomial Expression

Multinomial Algebraic Expression is an algebraic expression that contains two-term.

Example:
3x³ y² + 7x²y – 5xy + 4, 4a² + 9b² – 5c² – 2d², l + 9m + 7n – 5p, etc. are some of the examples for Multinomial Algebraic Expression.

Polynomial Expression

Polynomial Expression is an algebraic expression that contains the power of variables with a non-negative integer.

Example:
2x² + 4x + 6 is a polynomial.
x² + 4/x is not a polynomial. Because 4/x is negative.

Find the quick links of different algebraic expression concepts below. Simply, click the required link and prepare that particular concept with clear examples.

• Addition of Algebraic Expressions
• Subtraction of Algebraic Expressions
• Multiplication of Algebraic Expression
• Division of Algebraic Expressions

Algebraic Expression Examples

1. Express the following algebraic expressions with the help of signs and symbols.

(i) The sum of a and b
(ii) The subtraction of x from y.
(iii) The product of c and d.
(iv) x divided by 6.
(v) 5 divided by m.
(vi) The sum of 4 and p.
(vii) The product of z and 12.
(viii) 4 less than 6 times x.
(ix) Half of the product of 5 and x.
(x) One-tenth of x.
(xi) 4 less than the sum of m and n.
(xii) The values of c and d are equal.
(xiii) The values of a is greater than of b.
(xiv) 7 is less than y.

Solution:

(i) The sum of a and b
a + b
(ii) The subtraction of x from y.
y – x
(iii) The product of c and d.
cd
(iv) x divided by 6.
x/6
(v) 5 divided by m.
5/m
(vi) The sum of 4 and p.
4 + p
(vii) The product of z and 12.
12 × z
(viii) 4 less than 6 times x.
6x – 4
(ix) Half of the product of 5 and x.
5x/2
(x) One-tenth of x.
x/10
(xi) 4 less than the sum of m and n.
(m + n) – 4
(xii) The values of c and d are equal.
c = d
(xiii) The values of a is greater than of b.
a > b
(xiv) 7 is less than y.
7 < y

2. Express the following algebraic expressions in words

(i) c + d
(ii) 4a
(iii) a/6
(iv) a + b + 4
(v) 2m + n.
(vi) x + 3n
(vii) b – 5d
(viii) 3l – m
(ix) (m + 4n)/3
(x) s/3 + 9
(xi) 7 > 3x
(xii) a + b < 10

Solution:
(i) c + d
The sum of c and d
(ii) 4a
4 times of a
(iii) a/6
1/6 the part of a.
(iv) a + b + 4
The sum of a, b and 4
(v) 2m + n.
The sum of n and two times of m
(vi) x + 3n
The sum of x and three times of n
(vii) b – 5d
Deduction of 5 times of d from b
(viii) 3l – m
Deduction of m from 3 times of l
(ix) (m + 4n)/3
1/3 of the sum of m and four times n
(x) s/3 + 9
Sum of 1/3 rd portion of s and 9
(xi) 7 > 3x
7 is greater than three times of x
(xii) a + b < 10
Sum of a + b is less than 10

3. Express the following algebraic expressions using symbol if it is necessary.

(i) Sam has $14, Arun has $a more. How many dollars does Arun possess?
(ii) You worked out m sums yesterday. Today you have worked out 8 sums less. How many sums have you worked out today?
(iii) A car driver had earned L dollar on a day and $5 less on the next day. How much money has he earned on the next day?
(iv) Kyle has 7 pens. His father bought b more pens for her. How many pens now Kyle has?
(v) Ben had 15 chocolates, he lost x chocolates. How many chocolates are now remaining with him?
(vi) Anil is S years older than Arun. The present age of Arun is R years. How old is Anil now? What will be their ages after 5 years?
(vii) A painter earns $m daily. How much will he earn in 4 days?
(viii) There are C rows of trees in Akhil’s garden. In each row, there are 5 trees. How many trees are there in the garden?
(ix) You have two pencils. Your father gave you some more pencils? How many pencils are there with you now?

Solution:

(i) Sam has $14, Arun has $a more. How many dollars does Arun possess?
14 + a
(ii) You worked out m sums yesterday. Today you have worked out 8 sums less. How many sums have you worked out today?
m – 8
(iii) A car driver had earned L dollar on a day and $5 less on the next day. How much money has he earned on the next day?
L – 5
(iv) Kyle has 7 pens. His father bought b more pens for her. How many pens now Kyle have?
7 + b
(v) Ben had 15 chocolates, he lost x chocolates. How many chocolates are now remaining with him?
15 – x
(vi) Anil is S years older than Arun. The present age of Arun is R years. How old is Anil now? What will be their ages after 5 years?
Anil = S + R
Arun = R + 5
Anil = S + R + 5
(vii) A painter earns $m daily. How much will he earn in 4 days?
4m
(viii) There are C rows of trees in Akhil’s garden. In each row, there are 5 trees. How many trees are there in the garden?
5C
(ix) You have two pencils. Your father gave you some more pencils? How many pencils are there with you now?
2 + x

4. Write the algebraic expressions using symbols for the given problems?

(i) Monal had 5 color pens. She has lost some of them. How many color pencils she has now?
(ii) Sam’s age is 16 years.
(i) What was her age a years before?
(ii) What will be her age b years hence?
(iii) Five less than one-fourth of x
(iv) One-fifth of x.
(v) William is 5 years older than his brother Sonu. If Sonu’s age is m years, what will be William’s age?
(vi) The price of a dozen bananas is $ n. What will be the price of 6 dozen bananas?
(vii) The difference between the two numbers is L, the greater number is 20. Find a smaller number?
(viii) The product of two numbers is 25. One of them is c. Find the other?
(ix) Your age is 16 years now. What was your age h year ago? What will be your age after h years?

Solution:

(i) Monal had 5 color pens. She has lost some of them. How many color pencils she has now?
5 – x
(ii) Sam’s age is 16 years.
(i) What was her age a years before?
(ii) What will be her age b years hence?
(i) (16 – a) years
(ii) (16 + a) years
(iii) Five less than one-fourth of x
x/4 – 5
(iv) One-fifth of x.
x/5
(v) William is 5 years older than his brother Sonu. If Sonu’s age is m years, what will be William’s age?
(m + 5) years
(vi) The price of a dozen bananas is $ n. What will be the price of 6 dozen bananas?
6n
(vii) The difference between the two numbers is L, the greater number is 20. Find a smaller number?
Smaller number = 20 – L
(viii) The product of two numbers is 25. One of them is c. Find the other?
Other number = 25/c
(ix) Your age is 16 years now. What was your age h year ago? What will be your age after h years?
Age before h years = (16 – y) years
Age after h years = (16 + y) years

Simplification of algebraic fractions is nothing but finding the factors of both numerator and denominator and canceling the like terms. A fraction is a real number that represents the part of a single object from a group of objects. It is a combination of the numerator, denominator, and separator (/). The denominator of a fraction can not be zero. We can represent fractions as multiple fractions without changing the original one simply by altering its numerator, denominator values. Get the two simple methods to simply the fractions and solved examples in the below sections.

Algebraic Fractions

Fractions which has polynomial expressions in the numerator, denominator are called algebraic fractions. The algebraic fractions denominator can never be a zero. Every polynomial may be represented as an algebraic fraction with a denominator.

Adding Fractions:

To add two or more fractions, you must have a common denominator. To make a common denominator, just find the lcm of denominators or find the product of denominators. After making the denominator common, just add the numerators to get the addition of fractions.

Example: Solve 5 / 6 + 1 / 12?

LCM of 6, 12 is 12.

(5 * 2) / (6 * 2) + 1 / 12 = 10 / 12 + 1 / 12

= (10 + 1) / 12 = 11 / 12

Subtracting Fractions:

Addition and subtrcation of fractions is same. Here also, make a common denominator and subtract the numerator values.

Example: Solve [(x + 3) / (x – 3)] – (3 / (x – 3))

(x + 3 – 3) / (x – 3) = x / x – 3.

Multiplying Fractions:

Multiply numerators of the fractions and denominators of the fractions separately. Write the obtained numerator and denominator as a fraction to get the multiplication of fractions.

Example: Evaluate 1 / 2 * 23 / 25?

(1 * 23) / (2 * 25) = 23 / 50.

Dividing Fractions:

When you divide two fractions (a/b) / (c/d), you will get the answer as (ad / bc).

Example: Find 1/2 / 2/1?

(1 * 1) / (2 * 2) = 1 / 4.

Equivalent Fractions:

We can say that two or more fractions are equivalent when their numerator, denominators are equal. To make the fractions equal, you can divide or multiply the fractions by the same value.

Example: Check whether 12/42 and 2/7 are equivalent or not?

To make the numerator, the denominator of a second fraction equal to the first fraction, multiply the second fraction with 6.

(2 * 6) / (7 * 6) = 12 / 42

So, both are equivalent fractions.

What is Meant by Simplification of Algebraic Fractions?

Simplification of algebraic fractions means reducing a fraction to its lowest value. Both the numerator and denominator of a fraction are reduced so that it should have no common factor between them. The original value of the fraction will never change after the simplification. Therefore, it also called the equivalent fractions where numerator, denominator have no common factor except 1. An example is 25 / 15 = (5 * 5) / (5 * 3) = 5 / 3

Methods to Simplify Algebraic Fractions

Actually, we have two simple methods to simplify algebraic fractions. The one and only most important rule in the simplification of algebraic fractions is you must divide numerator, denominator with the same number at a time. Learn those steps in the below sections.

Method 1:

Simply, divide numerator and denominator by 2, 3, 4, 5… so on until you cannot find the common factor for those numbers.

Method 2:

• Find the highest common factor (HCF) for both the numerator, denominator.
• Divide both by the HCF number.
• The obtained result is a simplified fraction.

Whenever a polynomial expression is given in a fraction, then you can use this better and shorter way to simplify the fractions.

• Find the factors of both the numerator, denominator.
• Cancel the common factors in the numerator, denominator.
• Multiply the remaining values to get the result.

Example Questions on Simplification of Algebraic Fractions

Example 1.

Simplify the algebraic fraction: (10x³y³z²) / (2xy²z)

Solution:

Given fraction is (10x³y³z) / (2xy²z²)

The factors of the fraction are

(5 * 2 * x * x * x * y * y * y * z * z) / (2 * x * y * y * z)

We can see that ‘2’, ‘x’, ‘y * y’, ‘z’ are the common factors in the numerator and denominator. So, we cancel the common factors from the numerator and denominator.

= 5x²y / z

Example 2.

Simplify (2x – 10)⁶ / (x – 5)⁷.

Solution:

Given fraction is (2x – 10)⁶ / (x – 5)⁷

The common factors are (2 * (x – 5))⁶ / (x-5)⁷

= [2⁶ * (x-5) * (x-5) * (x-5) * (x-5) * (x-5) * (x-5)] / [(x-5) * (x-5) * (x-5) * (x-5) * (x-5) * (x-5) * (x-5)]

We can see that (x-5) is the common factor in the numerator, denominator. So, we cancel the common factors.

= 2⁶ / (x-5) = 64 / (x-5)

Example 3.

Reduce the algebraic fraction to its lowest term:

(x² + 2x – 35) / (x² – 25)

Solution:

Given fraction is (x² + 2x – 35) / (x² – 25)

x² + 2x – 35 = x² + 7x – 5x – 35 = x (x + 7) – 5(x + 7)

= (x – 5) (x + 7)

x² – 25 = (x – 5) (x + 5)

The common factors of the fraction is

[(x – 5) (x + 7)] / [(x – 5) (x + 5)]

Cancel the like terms

(x + 7) / (x + 5)

Example 4.

Simplify by adding and subtracting algebraic fractions:

[(2x – 1) / 3] – [(x – 5) / 6] + [(x – 4) / 2]

Solution:

Given that,

[(2x – 1) / 3] – [(x – 5) / 6] + [(x – 4) / 2]

The lcm of 3 , 6, 2 is 6.

Make the denominator as 6 for all parts of the expression.

[(2x – 1) / 3] – [(x – 5) / 6] + [(x – 4) / 2] = [2(2x – 1) / 6] – [(x – 5) / 6] + [3(x – 4) / 6]

= [(4x – 2) / 6] – [(x – 5) / 6] + [(3x – 12) / 6]

As, all denominators are equal perform arithmetic operations on numerator.

= [(4x – 2) – (x – 5) + (3x – 12)] / 6

= [4x – 2 – x + 5 + 3x – 12] / 6

= [6x – 9] / 6

= 3(x – 3) / 6

= 3(x – 3) / (3 * 2) = (x – 3) / 2

Example 5.

Simplify the following:

(i) [(x²y² + 3xy) / (4x² – 1)] / [(xy + 3) / (2x + 4)]

(ii) [(a² – 4b²) / (ab + 2b²)] * [(2b) / (a – 2b)]

Solution:

(i) Given that,

[(x²y² + 3xy) / (4x² – 16)] / [(xy + 3) / (2x + 4)]

(a/b) / (c/d) = (ad / bc)

[(x²y² + 3xy) / (4x² – 16)] / [(xy + 3) / (2x + 4)] = [(x²y² + 3xy) * (2x + 4)] / [(4x² – 16) * (xy + 3)]

The common factors are

= [(xy (xy + 3) * (2x + 4)] / [(2x -4) * (2x + 4) * (xy + 3)]

Cancel the common factors in both numerator, denominator.

= (xy) / (2x + 4)

= (xy) / [2(x + 2)]

(ii) Given that,

[(a² – 4b²) / (ab + 2b²)] * [(2b) / (a – 2b)]

Multiply numerator, denominator

= [(a² – 4b²) * (2b)] / [(ab + 2b²) * (a – 2b)]

The common factors are

= [(a + 2b) * (a – 2b) * 2b] / [(b(a + 2b) * (a – 2b)]

Cancel the like terms in both numerator, denominator

= (2b) / b = 2

FAQs on Simplification of Algebraic Fractions

1. How do you simplify an algebraic fraction?

In order to simplify an algebraic fraction, find the common factors for the numerator, denominator. Cancel the like terms in the numerator and denominator to get the simplified form.

2. How do you simplify algebraic fractions with quadratics?

Get the factors of polynomial expressions by using factorization. Find the common factors and cancel them. Perform the required operations to get the simplified fraction.

Arithmetic Fraction is a fraction which is in the form of a / b. Algebraic fractions contain polynomials either in the numerator or denominator. Get the definition of Arithmetic Fraction and Algebraic Fraction, some solved example questions in the further sections of this page.

Arithmetic Fraction Definition

Arithmetic fractions are represented as p / q. Where p is the numerator and q is the denominator which is not equal to zero. A number or expression in the form of a numerator/denominator is called the arithmetic fraction. It can also be expressed as p ÷ q. Some examples are 1/2, 3/10, 5/4, etc

When the numerator, denominator of arithmetic fractions are multiplied or divided by the same quantity, then the value of the fraction remains the same. Arithmetic fractions are mostly monomial quantities or they can be reduced to monomials.

Some Examples of Arithmetic Fraction

• 12/16 is an arithmetic fraction having numerator 12 and denominator 16.
• 17/85 is also an arithmetic fraction having numerator 17 and denominator 85.
• 18/81 = 2/9
• 16/32 = 1/2

Algebraic Fraction Definition

An algebraic Fraction is also a fraction whose numerator, denominators are algebraic expressions. Both numerator and denominator are polynomials.

Examples of Algebraic Fraction

• If both denominator and numerator are monomials.
  • a/b, x/y, ax²/bc, my²/n, etc
• If the denominator is monomial and the numerator is polynomial or binomial.
  • x+y / z, a²+bc+b² / ab, ab+bc+ca / a, etc
• If the denominator is binomial/polynomial and the numerator is monomial.
  • a/b+c, x/y-z, a/ab+bc+ca, etc
• If numerator and denominator are binomial or polynomial.
  • x+y/x-y, a+b+c/a+c, a²+2bc+c/a+b, etc

A fraction represents equal parts of a collection or whole. Algebraic fraction means a fraction whose numerator or denominator is a polynomial expression. Check the examples of algebraic fractions, its definition, and solved example questions. You can also see how to perform addition, multiplication, subtraction, and division between algebraic fractions.

Algebraic Fraction Definition

Fractions that have a polynomial expression in the numerator and denominator are called algebraic fractions. Denominators of the algebraic fractions can never be zero. You can write every polynomial as an algebraic fraction with a denominator.

Examples of Algebraic Fractions

• (x² + 2x + 3) / 3 is an algebraic fraction with integral denominator 3.
• (x + 3) / 5 is an algebraic fraction with integral denominator 5.
• 6 / (a + 5b + 3) is an algebraic fraction with integral numerator 6.
• 2 / (x + 3y) is an algebraic fraction with integral numerator 2.
• (x + y) / (x² + 10x + 7) is an algebraic fraction with numerator as linear polynomial and denominator as a quadratic polynomial.
• (y² – 11y + 32) / (y + 2) is an algebraic fraction with numerator as quadratic polynomial and denominator as a linear polynomial.

Operations with Algebraic Fractions

Below mentioned are the various operations that can be performed on algebraic fractions.

1. Reducing Fraction

To reduce the algebraic fraction, first, find the factors of numerator and denominator. Later cancel the common factors.

Example: (12x³y²) / (28xy)

(12x³y²) / (28xy) = (4 * 3 * x * x * x * y * y) / (7 * 4 * x * y)

= (3 * x * x * y) / 7 = (3x²y) / 7 = (3/7) x²y

2. Multiplying Algebraic Fractions

While multiplying 2 fractions, get the factors of the fraction and then reduce it to the lowest terms. Then multiply the numerators together, and denominators together to check the result.

Example: [(x + 1) / (5y + 10)] * [(y+2) / (x² + 2x + 1)]

= [(x + 1) / ((5(y + 2))] * [(y+2) / (x² + x + x + 1)]

= [(x + 1) / (5] * [1 / (x (x + 1) + 1 (x + 1)] = [(x + 1) / (5] * [1 / ((x + 1) (x + 1)]

= (1/5) * (1 / (x + 1)) = 1 / (5(x + 1))

3. Adding Fractions

You must have a common denominator to add two fractions. Simply, get the lcm or multiply those two denominators. Then, add the numerators.

Example: [(x – 4) / (x + 1)] + [3 / (x + 2)]

= [(x – 4) (x + 2) / (x + 1) (x + 2)] + [(3(x + 1)) / (x + 1) (x + 2)]

= [(x² – 4x + 2x – 8) / (x + 1) (x + 2)] + [(3x + 3) / (x + 1) (x + 2)]

= [(x² – 2x – 8) / (x + 1) (x + 2)] + [(3x + 3) / (x + 1) (x + 2)] = [(x² – 2x – 8 + 3x + 3) / (x + 1) (x + 2)]

= (x² + x – 5) / [(x + 1) (x + 2)]

4. Subtracting fractions

To subtract fractions, make a common denominator by finding the LCD and changing each fraction to an equivalent fraction. Then subtract those fractions.

Example: (2 / x) – (3 / y)

= (2y / xy) – (3x / xy) = (2y – 3x) / xy

5. Dividing Fractions

To divide algebraic fractions, invert the second fraction and multiply.

Example: (2 / x²) / (5 / x)

= (2 / x²) * (x / 5) = 2x / 5x² = 2 / 5x

Example Questions

Example 1. 

Solve [5 / (x + 6)] + [10 / (x + 1)]

Solution:

Given fraction is [5 / (x + 6)] + [10 / (x + 1)]

Both fractions denominators are different. So find the lcm

LCM of (x + 6) and (x + 1) is (x + 6) * (x + 1)

[5 / (x + 6)] + [10 / (x + 1)] = [(5(x + 1)) / (x + 6) (x + 1)] + [(10(x + 6)) / (x + 6) (x + 1)]

= [(5x + 5) / (x + 6) (x + 1)] + [(10x + 60) / (x + 6) (x + 1)]

= [(5x + 5 + 10x + 60) / (x + 6) (x + 1)] = [(15x + 65) / (x + 6) (x + 1)]

[5 / (x + 6)] + [10 / (x + 1)] = [5(3x + 13) / (x + 6) (x + 1)]

Example 2.

Solve [(x² + 13x + 35) / (x + 4)] / [(x² – 3x – 40) / (x – 6)]

Solution:

Given fraction is [(x² + 13x + 35) / (x + 4)] / [(x² – 3x – 40) / (x – 6)]

[(x² + 13x + 35) / (x + 4)] / [(x² – 3x – 40) / (x – 6)] = [(x² + 13x + 35) / (x + 4)] * [(x – 6) / (x² – 3x – 40)]

= [(x² + 7x + 5x + 35) / (x + 4)] * [(x – 6) / (x² – 8x + 5x – 40)] = [(x(x + 7) + 5 (x + 7) / (x + 4)] * [(x – 6) / (x(x – 8) + 5(x – 8)]

= [((x + 7) (x + 5) / (x + 4)] * [(x – 6) / ((x – 8) (x + 5)]

= [(x + 7) (x + 5) (x – 6)] / [(x + 4) (x – 8) (x + 5)] = [(x + 7) (x – 6)] / [(x + 4) (x – 8)]

Example 3.

Perform the indicated operation.

(i) [10 / y] – [15 / y² – 10y + 25]

(ii) [x / (x² + 5x + 6)] – [2 / (x² + 3x + 2)]

Solution:

(i) Given fraction is [10 / y] – [15 / y² – 10y + 25]

= [10 / y] – [15 / (y² – 5y – 5y + 25]

= [10 / y] – [15 / (y(y – 5) – 5(y – 5)] = [10 / y] – [15 / (y – 5)(y – 5)]

= [10 (y – 5)² / y(y – 5)²] – [15y / y(y – 5)²]

= (10 (y – 5)² – 15y) / y(y – 5)² = 10(y² – 10y + 25) / y(y – 5)²

= [10(y² + 25) – 100y – 15y] / [y(y – 5)²] = [10(y² + 25) – 115y] / [y(y – 5)²]

(ii) Given fraction is [x / (x² + 5x + 6)] – [2 / (x² + 3x + 2)]

= [x / (x² + 3x + 2x + 6)] – [2 / (x² + 2x + x + 2)]

= [x / (x(x + 3) + 2(x + 3)] – [2 / (x(x + 2) + 1(x + 2)]

= [x / ((x + 3)(x + 2)] – [2 / (x + 2)(x + 1)] = [x(x + 1) / (x + 2)(x + 1)(x + 3)] – [2(x + 3) / (x + 2)(x + 1)(x + 3)]

= [x² + x – 2x – 6] / [(x + 2)(x + 1)(x + 3)] = [x² – x – 6] / [(x + 2)(x + 1)(x + 3)]

= [(x – 3) (x + 2)] / [(x + 2)(x + 1)(x + 3)]

= [x – 3] / [(x + 1)(x + 3)]

Calculating Multiplication of Algebraic Expression is not that easy as you think. Students must follow some rules to find Multiplication of Algebraic Expression. If students to do any small mistakes, then it may lead to wrong answers. So, every operation is important while finding Multiplication of Algebraic Expression. Different types of problems and methods to solve problems are given in this article clearly. Students can easily understand the method of solving the Multiplication of Algebraic Expression after reading this article completely.

Rules to Find Multiplication of Algebraic Expression

1. Product of two same signs is positive, and also the product of different signs is negative.
2. If a is a variable and x, y are two positive integers, then (aᵐ × aⁿ) = a^(m+n)

Types of Algebraic Expression Multiplication

There are different types of multiplication occurs while finding Algebraic Expressions
1) Multiplication of Two Monomials
2) Multiplication of a Polynomial by a Monomial
3) Multiplication of Two Binomials
4) Multiplication by Polynomial

How to Find Multiplication of Two Monomials?

1. Write the two numbers along with the multiplication sign
2. Multiply the numbers.
3. If you find the bases are the same then add the exponents.

Product of two monomials = (Multiplication of their numerical coefficients) × (Multiplication of their variable parts)

Solved Examples

1. Find the product of 4xy and -6x²y³

Solution:
Given that 4xy and -6x²y³
4xy × -6x²y³
Multiply the coefficient. If the signs are the same, then the resultant coefficient is positive. Or else, if the signs are not the same, then the resultant coefficient is negative.
4 × -6 = -24
Multiply the variables. If the base of the variables is the same, then add the powers.
xy × x²y³
x^(1 + 2)y^(1 + 3)
x³y⁴
Multiply coefficient and variables.
-24x³y⁴

The required answer is -24x³y⁴

2. Find the product of 7ab², -4a²b, and -5abc?

Given that 4ab², -6a²b, and -7abc
4ab² × -6a²b × -7abc
Multiply the coefficient. If the signs are the same, then the resultant coefficient is positive. Or else, if the signs are not the same, then the resultant coefficient is negative.
4 × -6 × -7 = 168
Multiply the variables. If the base of the variables is the same, then add the powers.
ab² × a²b × abc
a^(1 + 2 + 1)b^(2 + 1 + 1)c
a⁴b⁴c
Multiply coefficient and variables.
168a⁴b⁴c

The required answer is 168a⁴b⁴c

3. Find the product of 5ab and 3a³b²

Given that 5ab and 3a³b²
5ab × 3a³b²
Multiply the coefficient. If the signs are the same, then the resultant coefficient is positive. Or else, if the signs are not the same, then the resultant coefficient is negative.
5 × 3 = 15
Multiply the variables. If the base of the variables is the same, then add the powers.
ab × a³b²
a^(1 + 3)b^(1 + 2)
a⁴b³
Multiply coefficient and variables.
15a⁴b³

The required answer is 15a⁴b³

4. Find the product of 6x²y, 9z²x, and -6xy²z?

Given that 6x²y, 9z²x, and -6xy²z
6x²y × 9z²x × -6xy²z
Multiply the coefficient. If the signs are the same, then the resultant coefficient is positive. Or else, if the signs are not the same, then the resultant coefficient is negative.
6 × 9 × -6 = -324
Multiply the variables. If the base of the variables is the same, then add the powers.
x²y × z²x × xy²z
x^(2 + 1 + 1)y^(1 + 3)z^(2 + 1)
x⁴y⁴z³
Multiply coefficient and variables.
-324x⁴y⁴z³

The required answer is -324x⁴y⁴z³

How to Find Multiplication of a Polynomial by a Monomial?

1. Use a distributive law and multiply a polynomial by a monomial.
2. Multiply each individual term in the parenthesis by a monomial.

Multiply each term of the polynomial by the monomial, using the distributive law a × (b + c) = a × b + a × c.

Solved Examples

1. 6a²b² × (2a² – 6ab + 8b²)

Solution:
Given that 6a²b² × (2a² – 6ab + 8b²)
Apply distributive law of multiplication for the given terms.
a × (b + c) = a × b + a × c.
6a²b² × (2a² – 6ab + 8b²) = (6a²b²) × (2a²) + (6a²b²) × (-6ab) + (6a²b²) × (8b²)
Find the final expression by multiplying each term.
12a⁴b² – 36a³b³ + 48a²b⁴.

The required expression is 12a⁴b² – 36a³b³ + 48a²b⁴.

2. (-9x²y) × (2x²y – 5xy² + x – 7y)

Solution:
Given that (-9x²y) × (2x²y – 5xy² + x – 7y)
Apply distributive law of multiplication for the given terms.
a × (b + c) = a × b + a × c.
(-9x²y) × (2x²y – 5xy² + x – 7y) = (-9x²y) × (2x²y) + (-9x²y) × (-5xy²) + (-9x²y) × (x) + (-9x²y) × (-7y)
Find the final expression by multiplying each term.
-18x⁴y² + 45x³y³ – 9x³y + 63x²y²

The required expression is -18x⁴y² + 45x³y³ – 9x³y + 63x²y².

3. 0(x^4 + 2x^3 + 3x^2 + 9x + 1)

Solution:
Given that 0(x^4 + 2x^3 + 3x^2 + 9x + 1)
Any polynomial multiply by zero is zero.
Therefore, multiplying (x^4 + 2x^3 + 3x^2 + 9x + 1) with 0 is 0.

The answer is 0.

4. 1 (5 x^4 – 8 )

Solution:
Given that 1 (5 x^4 – 8 )
Any polynomial multiply by 1 is the polynomial itself.
Therefore, multiplying (5 x^4 – 8 ) with 1 is (5 x^4 – 8).

The required expression is (5 x^4 – 8).

3. How to Find Multiplication of Two Binomials?

Use two methods to find the Multiplication of Two Binomials. Students can use a horizontal method or Column wise multiplication to find Multiplication of Two Binomials.

How to Find Multiplication of Two Binomials using Horizontal method?

1. Firstly, note down two binomials.
2. Apply the distributive law of multiplication over addition twice.
3. Find the final expression of multiplication.

How to Find Multiplication of Two Binomials using Column wise multiplication?

1. Write one binomial expression under another expression.
2. Multiply the first binomial expression with the first term of the second binomial expression.
3. Multiply the first binomial expression with the second term of the second binomial expression.
4. Note down the first resultant expression and write the second resultant expression below the first resultant expression with like terms comes at the same column.
5. Add the first and second expressions to get the final expression.

Solved Examples

1. Multiply (m + n) × (r + s)

Solution:

Horizontal Method:
Note down two binomials.
(m + n) × (r + s)
Apply the distributive law of multiplication over addition twice.
m × (r + s) + n × (r + s)
(m × r + m × s) + (n × r + n × s)
mr + ms + nr + ns

The required expression is mr + ms + nr + ns.

Column wise multiplication

Write one binomial expression under another expression
m + n
× (r + s)
—————————-
rm + rn —> Multiplication of r with (m + n)
+ ms + ns —> Multiplication of s with (m + n)
—————————-
mr + ms + nr + ns

The required expression is mr + ms + nr + ns.

2. Multiply (2x + 4y) and (4x – 6y)

Solution:

Horizontal Method:
Note down two binomials.
(2x + 4y) and (4x – 6y)
Apply the distributive law of multiplication over addition twice.
2x × (4x – 6y) + 4y × (4x – 6y)
(2x × 4x – 2x × 6y) + (4y × 4x – 4y × 6y)
(8x² – 12xy) + (16xy – 24y²)
8x² – 12xy + 16xy – 24y²
8x² + 4xy – 24y².

The required expression is 8x² + 4xy – 24y².

Column wise multiplication
Write one binomial expression under another expression
(2x + 4y)
× (4x – 6y)
—————————-
8x² + 16xy —> Multiplication of 4x with (4x – 6y)
– 12xy – 24y² —> Multiplication of 4y with (4x – 6y)
—————————-
8x² + 4xy – 24y².

The required expression is 8x² + 4xy – 24y².

3. Multiply (4x² + 2y²) by (3x² + 5y²)

Solution:

Horizontal Method:
Note down two binomials.
(4x² + 2y²) × (3x² + 5y²)
Apply the distributive law of multiplication over addition twice.
4x² (3x² + 5y²) + 2y² (3x² + 5y²)
(12x⁴ + 20x²y²) + (6x²y² + 10y⁴)
12x⁴ + 20x²y² + 6x²y² + 10y⁴
12x⁴ + 26x²y² + 10y⁴

The required expression is 12x⁴ + 26x²y² + 10y⁴

Column wise multiplication
Write one binomial expression under another expression
(4x² + 2y²)
× (3x² + 5y²)
—————————-
12x⁴ + 20x²y² —> Multiplication of 3x² with (4x² + 2y²)
6x²y² + 10y⁴ —> Multiplication of 5y² with (4x² + 2y²)
—————————-
12x⁴ + 26x²y² + 10y⁴

The required expression is 12x⁴ + 26x²y² + 10y⁴

4. How to Find Multiplication by Polynomial?

Apply horizontal method or column multiplication to find Multiplication by Polynomial.

Solved Examples:

1. Multiply (6x² – 4x + 9) by (3x – 7)

Solution:

Horizontal Method:
Given that (6x² – 4x + 9) by (3x – 7)
(3x – 7) × (6x² – 4x + 9)
3x (6x² – 4x + 9) – 7 (6x² – 4x + 9)
(18x³ – 12x² + 27x ) + (- 42x² + 28x – 63)
18x³ – 12x² + 27x – 42x² + 28x – 63
18x³ – 54x² + 55x – 63

The required expression is 18x³ – 54x² + 55x – 63

Column wise multiplication
Write one binomial expression under another expression
(6x² – 4x + 9)
× (3x – 7)
—————————-
18x³ – 12x² + 27x —> Multiplication of 3x with (6x² – 4x + 9)
– 42x² + 28x – 63 —> Multiplication of -7 with (6x² – 4x + 9)
—————————-
18x³ – 54x² + 55x – 63

The required expression is 18x³ – 54x² + 55x – 63

2. Multiply (3x² – 6x + 2) by (2x² + 9x – 5)

Solution:

Horizontal Method:
Given that (3x² – 6x + 2) by (2x² + 9x – 5)
2x² (3x² – 6x + 2) + 9x (3x² – 6x + 2) – 5 (3x² – 6x + 2)
(6x⁴ – 12x³ + 4x²) + ( + 27x³ – 54x² + 18x ) + (- 15x² + 30x – 10 )
6x⁴ – 12x³ + 4x² + 27x³ – 54x² + 18x – 15x² + 30x – 10
6x⁴ + 15x³ – 65x² + 48x – 10

The required expression is 6x⁴ + 15x³ – 65x² + 48x – 10

Column wise multiplication
Write one binomial expression under another expression
(3x² – 6x + 2)
× (2x² + 9x – 5)
—————————-
6x⁴ – 12x³ + 4x² —> Multiplication of 2x² with (6x² – 4x + 9)
+ 27x³ – 54x² + 18x —> Multiplication of 9x with (6x² – 4x + 9)
– 15x² + 30x – 10 —> Multiplication of – 5 with (6x² – 4x + 9)
—————————-
6x⁴ + 15x³ – 65x² + 48x – 10

The required expression is 6x⁴ + 15x³ – 65x² + 48x – 10

3. Multiply (3x³ – 4x² – 2x + 9) by (5 – 6x + 7x²)

Solution:

Horizontal Method:
Given that (3x³ – 4x² – 2x + 9) by (5 – 6x + 7x²)
5 (3x³ – 4x² – 2x + 9) – 6x (3x³ – 4x² – 2x + 9) + 7x² (3x³ – 4x² – 2x + 9)
(15x³ – 20x² – 10x + 45) + (-18x⁴ + 24x³ + 12x² – 54x ) + (21x⁵ – 28x⁴ – 14x³ + 63x²)
15x³ – 20x² – 10x + 45 -18x⁴ + 24x³ + 12x² – 54x + 21x⁵ – 28x⁴ – 14x³ + 63x²
21x⁵ -46x⁴ + 25x³ + 55x² – 10x + 45

The required expression is 21x⁵ -46x⁴ + 25x³ + 55x² – 10x + 45

Column wise multiplication
Write one binomial expression under another expression
(3x³ – 4x² – 2x + 9)
× (5 – 6x + 7x²)
—————————-
15x³ – 20x² – 10x + 45 —> Multiplication of 5 with (3x³ – 4x² – 2x + 9)
-18x⁴ + 24x³ + 12x² – 54x —> Multiplication of – 6x with (3x³ – 4x² – 2x + 9)
21x⁵ – 28x⁴ – 14x³ + 63x² —> Multiplication of 7x² with (3x³ – 4x² – 2x + 9)
—————————-
21x⁵ -46x⁴ + 25x³ + 55x² – 10x + 45

The required expression is 21x⁵ -46x⁴ + 25x³ + 55x² – 10x + 45