Algebraic Expressions and Identities Chapter Notes | Mathematics (Maths) Class 8 PDF Download

What are Algebraic Expressions?

Any mathematical expression that consists of numbers, variables and operations are called Algebraic Expression. 

Components of Algebraic Expression

• Constants: Fixed numerical values. For example, in the expression $3x+43x\; +\; 4$3$x$ +4, the number 4 is a constant.
• Variables: Symbols that represent unknown values and can change. For example, $xx$ in $3x+43x\; +\; 4$3$x$ +4 is a variable.
• Coefficients: Numbers that are multiplied by the variables. In the expression $3x3x$, 3 is the coefficient of $xx$.
• Operators: Symbols that indicate mathematical operations, such as $+$+, $--$, $\backslash times$×, and $\backslash div$÷.

Monomials, Binomials and Polynomials

Addition and Subtraction of Algebraic Expressions

Steps to Add or Subtract Algebraic Expression

• First of all, we have to write the algebraic expressions in different rows in such a way that the like terms come in the same column.
• Add them as we add other numbers.
• If any term of the same variable is not there in another expression then write is as it is in the solution.

Example 1:Add 15p² – 4p + 5 and 9p – 11

Solution:

Write down the expressions in separate rows with like terms in the same column and add. 

Example 2: Subtract 5a² – 4b² + 6b – 3 from 7a² – 4ab + 8b² + 5a – 3b.

Solution:

For subtraction also write the expressions in different rows. But to subtract we have to change their signs from negative to positive and vice versa.

Multiplication of Algebraic Expression: Introduction

While multiplying, we need to take care of some points about the multiplication of like and unlike terms.

1. Multiplication of Like Terms

Example3:The product of 4x and 3x will be 12x².
(4x) x (3x) = 12x²

Example 4:The product of 5x, 3x and 4x will be 60x³.
(5x) x (3x) x (4x) = 60x³

2. Multiplication of Unlike Terms

• The coefficients will get multiplied.
• The power will remain the same if the variable is different.
• If some of the variables are the same then their powers will be added.

Example5:The product of 2p and 3q will be 6pq
(2p) x (3q) = 6pq

Example 6:The product of 2x²y, 3x and 9 will  be 54x³y
(2x²y) x (3x) x (9) = 54x³y

Multiplying a Monomial by a Monomial

Multiplying Two Monomials

While multiplying two polynomials the resultant variable will come by

• The coefficient of product = Coefficient of the first monomial × Coefficient of the second monomial
• The algebraic factor of product = Algebraic factor of the first monomial × Algebraic factor of the second monomial.

Example 7: 

(3x²) × (4x³) = 12x²⁺³ = 12x⁵
Explanation: Multiply the coefficients 3 × 4 = 12 and add the exponents of x (2 + 3 = 5).

Example 8:

(2a³b) × (5ab²) = 10a³⁺¹b¹⁺² = 10a⁴b³

Explanation: Multiply the coefficients 2 × 5 = 10, add the exponents of a (3 + 1 = 4), and add the exponents of b (1 + 2 = 3).

Multiplying Three or More Monomials

While multiplying three or more monomial the criterion will remain the same.

Example 9:

4xy × 5x²y² × 6x³y³ = (4xy × 5x²y²) × 6x³y³

= 20x³y³ × 6x³y³

= 120x³y³ × x³y³

= 120(x³ × x³) × (y³ × y³)

= 120x⁶y⁶

= 120x⁶y⁶

We can do it in another way also:

4xy × 5x²y² × 6x³y³

= (4 × 5 × 6) × (x × x² × x³) × (y × y² × y³)

= 120x⁶y⁶

Example 10: Find the volume of each rectangular box with given length, breadth, and height.

Length	Breadth	Height
2ax	3by	5cz
m2n	n2p	p2m
2q	4q2	8q3

Volume = length × breadth × height

For (i) volume = (2ax) × (3by) × (5cz)
= 2 × 3 × 5 × (ax) × (by) × (cz)
= 30abcxyz

For (ii) volume = m²n × n²p × p²m
= (m² × m) × (n × n²) × (p × p²)
= m³n³p³

For (iii) volume = 2q × 4q² × 8q³
= 2 × 4 × 8 × q × q² × q³
= 64q⁶

Multiplying a Monomial by a Polynomial

Multiplying a monomial by a binomial

To multiply a monomial with a binomial we have to multiply the monomial with each term of the binomial.

Example 11: Simplify 2x × (3x + 5xy).
Solution: 2x × (3x + 5xy) = (2x × 3x) + (2x × 5xy)
= 6x² + 10x²y

Example 12: Simplify a² × (2ab − 5c).
Solution: a² × (2ab − 5c) = (a² × 2ab) − (a² × 5c)
= 2a³b − 5a²c

Multiplying a Monomial by a Trinomial

Consider 4q × (5q² + 6q + 8).
As in the earlier case, we use the distributive law:
4q × (5q² + 6q + 8) = (4q × 5q²) + (4q × 6q) + (4q × 8)
= 20q³ + 24q² + 32q
Multiply each term of the trinomial by the monomial and add products.

Example 13: Simplify the expressions and evaluate them as directed:

(i) x(x − 4) + 3 for x = 2

(ii) 5z(3z − 9) − 4(z − 2) − 45 for z = −1

Solution:
(i) x(x − 4) + 3 = x² − 4x + 3
For x = 2:
x² − 4x + 3 = (2)² − 4(2) + 3
= 4 − 8 + 3 = −1
(ii) 5z(3z − 9) − 4(z − 2) − 45 = 15z² − 45z − 4z + 8 − 45
= 15z² − 49z − 37
For z = −1:
15z² − 49z − 37 = 15(−1)² − 49(−1) − 37
= 15 + 49 − 37 = 27

Multiplying a Polynomial by a Polynomial

Multiplying a binomial by a binomial

We use the distributive law of multiplication in this case. Multiply each term of a binomial with every term of another binomial. After multiplying the polynomials we have to look for the like terms and combine them.

Example 14: Simplify (3a + 4b) × (2a + 3b)

Solution:

(3a + 4b) × (2a + 3b)

= 3a × (2a + 3b) + 4b × (2a + 3b)    [distributive law]

= (3a × 2a) + (3a × 3b) + (4b × 2a) + (4b × 3b)

= 6 a² + 9ab + 8ba + 12b²

= 6 a² + 17ab + 12b²     [Since ba = ab]

Multiplying a Binomial by a Trinomial

In this, we also have to multiply each term of the binomial with every term of the trinomial.

Example 15: Simplify (p + q) (2p – 3q + r) – (2p – 3q) r.

Solution:

We have a binomial (p + q) and one trinomial (2p – 3q + r)

(p + q) (2p – 3q + r)

= p × (2p – 3q + r) + q × (2p – 3q + r) [distributive law]

= 2p² – 3pq + pr + 2pq – 3q² + qr

= 2p² – pq – 3q² + qr + pr    (–3pq and 2pq are like terms)

(2p – 3q) r = 2pr – 3qr

Therefore,

(p + q) (2p – 3q + r) – (2p – 3q) r

= 2p² – pq – 3q² + qr + pr – (2pr – 3qr)

= 2p² – pq – 3q² + qr + pr – 2pr + 3qr

= 2p² – pq – 3q² + (qr + 3qr) + (pr – 2pr)

= 2p² – 3q² – pq + 4qr – pr

What is Identity? 

An identity is an equality that is true for every value of the variable but an equation is true for only some of the values of the variables.

So, an equation is not an identity.

Like, x² = 1, is valid if x is 1 but is not true if x is 2.so it is an equation but not an identity.

Standard Identities

(a + b)² = a² + 2ab + b²

(a - b)² = a² – 2ab + b²

a² – b² = (a + b) (a - b)

(x + a) (x + b) = x² + (a + b)x + ab

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

These identities are useful in carrying out squares and products of algebraic expressions. They give alternative methods to calculate products of numbers and so on.