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## Download RS Aggarwal Solutions Class 7 Maths Chapter 6 Ex 6.3 PDF

RS Aggarwal Solutions Class 7 Maths Chapter 6 Ex 6.3

## RS Aggarwal Solutions Class 7 Maths Chapter 6 Ex 6.3 – Overview

In this exercise, you will solve the problems related to the multiplication operation between the binomial and the monomial.

It can be done via associative property or distributive property. In contrast to this, when we multiply the binomial with another binomial, it also requires the use of the same rules. All of this can be done using the shortcut method of the FOIL i.e., ( First, Out, Inner, Last).

### Multiplying Monomials

### Multiplying a Monomial by a Binomial

Multiplying a monomial with a binomial is done using the distributive property rule. Let us take the binomial like 3x + 5 and multiply it further the monomial 2x.

### Multiplying Two Binomials

While multiplying two binomials, we use the distributive property two times.

### Using FOIL Method (First, Out, Inner, Last)

This is the shortcut method that can easily replace the multiplication operation done between monomials and binomials via distribution property.

- Initially multiply the first number contained in each of the binomials, the outside number contained in each of the binomials, the inside number contained in each of the binomials, and the last number contained in each of the binomial” – this is how we perform the calculation via the FOIL method.
- Accumulate all the like numbers, this is to be done by performing the addition operation between the outside terms. This way the result that will see is exactly going to be the same as the distributive property method.

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### How to multiply a monomial with a binomial?

Multiplying a monomial with a binomial is done using the distributive property rule. Let us take the binomial like 3x + 5 and multiply it further the monomial 2x.

(3x + 5) × 2x = 3x (2x) + 5 (2x)

(3x + 5) × 2x = 6x² + 10x

### How to multiply a binomial?

While multiplying two binomials, we use the distributive property two times.

Eg: ( x + 2 ) and ( x + 5 ), here, it can be solved via using the distributive property as mentioned below.

( x + 2 ) × ( x + 5 ) = x ( x + 5 ) + 2 ( x + 5 )

( x + 2 ) × ( x + 5 ) = x² + 5x + 2x + 10

( x + 2 ) × ( x + 5 ) = x² + 7x + 10