CBSE Notes For Class 8 Maths Chapter 8 Algebraic Expressions And Identities

1.0Introduction 

In earlier math classes, you might have learned how to add and subtract simple algebraic expressions. However, when it comes to slightly larger equations, properly arranging your equation, and the methods to write it down become more important. It also becomes necessary to understand the multiplication of such expressions. 

The concept of monomials, binomials, trinomials, and polynomials is a must-know basis to easily learn the multiplication of algebraic expressions. 

2.0CBSE Class 8 Math Chapter 8 Algebraic Expressions and Identities- Revision Notes

How to Add and Subtract algebraic expressions?

Let us say we have to add the two expressions,  8x2+3x-4 and 10x+2. To subtract them, arrange them as follows:       

$\begin{array}{r}8x^2+3x-4\\ -10x+2\\ 8x^2-7x-6\end{array}$

Here, it is important to note that in general math, (-4)-(+2) becomes (-4)-2, which is -6.

Multiplication of Monomials

Algebraic expressions that contain only one term are called monomials. 

1. Multiplying two monomials

• In simple terms, 5x=x+x+x+x+x=5x. 
• In the same way, 3(5x)=5x+5x+5x=15x.
• Take an expression $5xy\times 2x{y}^2=(5\times 2)\times (xy\times x{y}^2)$, which becomes $10x^2{y}^3$
• Here, the numbers are multiplied separately while the powers of variables are collected from the algebraic parts of both monomials.

2. Multiplying three or more monomials

• With more than two monomials, focus on multiplying the first two monomials. The solution from multiplying the first two is then multiplied with the third monomial.
• This method can fit as many monomials as required in your math problem.

Multiplication of Monomials and Polynomials

An expression with two terms is a binomial, three is a trinomial, and more than that is a polynomial.

1. Multiplying a monomial by a binomial:

• Say we have to multiply the monomial 3x with a binomial 2x+y.
• Here, we will have to use the distributive law for math, which makes it $3x\times (2x+y)=(3x\times 2x)+(3x\times y)=6x^2+3xy.$

2. Multiplying a monomial by a trinomial:

• Here, each term in the trinomial is multiplied by the monomial using distributive law. The products of this are then added together.
• Multiply 3x by $x^2+2x+4.$

$x\cdot (x^2+2x+4)=(3x\cdot x^2)+(3x\cdot 2x)+(3x\cdot 4)$

$=3x^3+6x^2+12x$

Multiplying Polynomials

1. Multiplying a binomial by a binomial

• This also follows the same distributive law. When multiplying polynomials, terms should always be identified and combined, if any.

2. Multiplying a binomial by a trinomial

• Each of the three terms in the trinomial has to be multiplied by each of the two terms in the binomial. 
• Everything else follows the same rules of distribution as before.

3.0Solving Questions Related to Algebraic Expressions and Identities

Question 1: Simplify $5xy\times 7x^2{y}^2\times 8x^2{y}^2\times 2xy$

Solution:

$=5xy\times 7x^2{y}^2\times 8x^2{y}^2\times 2xy=(5xy\times 7x^2{y}^2)(8x^2{y}^2\times 2xy)$

$=35(x^3{y}^3)\times 16(x^3{y}^3)$

$=560x^6{y}^6$

Question 2: Simplify (2x+3y)(3x+4y) 

Solution:

$=3x\times (2x+3y)+4y\times (2x+3y)$

$=(3x\times 2x)+(3x\times 3y)+(4y\times 2x)+(4y\times 3y)$

$=6x^2+9yx+8yx+12y^2$

$=6x^2+17xy+12y^2.$

Question 3: Verify that (3x + 5y)² – 30xy = 9x² + 25y²

Solution: Taking L.H.S. 

= (3x + 5y)² – 30xy 

Taking identity (a + b)² = a² + 2ab + b²

= (3x)² + 2 × 3x × 5y + (5y)² – 30xy 

= 9x² + 30xy + 25y² – 30xy 

= 9x² + 25y²

L.H.S = R.H.S 

Question 4: Verify that (11pq + 4q)² – (11pq – 4q)² = 176pq² 

Solution: 

L.H.S. 

= (11pq + 4q)² – (11pq – 4q)² 

= (11pq + 4q + 11pq – 4q) × (11pq + 4q – 11pq + 4q) 

[using a² – b² = (a – b) (a + b), here a = 11pq + 4q and b = 11 pq – 4q] 

= (22pq) (8q) = 176 pq²

L.H.S = R.H.S 

4.0Key Features of CBSE Maths Notes for Class 8 Chapter 8

• Notes are aligned with the latest CBSE math curriculum.
• Easy to understand, with simple examples.
• Crisp and short explanations for quick revision.
• Notes are checked thoroughly and are 100% accurate.

5.0Sample Question on CBSE Maths Notes for Class 8 Chapter 8

Q1: How can we substitute values in these expressions?

Take this expression

$3x\times (4x^2+5x+7)=(3x\times 4x^2)+(3x\times 5x)+(3x\times 7)=12x^3+5x^2+21x.$

If you are instructed to solve it x=1 , you can simply substitute 1 as the value of x to obtain: 12(1)+5(1)+21(1)=38.