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## All Questions

#### Give possible expressions for the length and breadth of the rectangle whose area is given by

Length (2a+3)

Or

Length = (2a-1)

Solution:

Given : Area of rectangle is 4a2+4a-3    …..(1)

Factorize equation 1, we get

=4a2+6a-2a-3

=2a(2a+3)-1(2a+3)

=(2a+3)(2a-1)

We know that area of rectangle is length × breadth

Hence

Length (2a+3)

Or

Length = (2a-1)

#### Find the value of(i) when (ii) when

(i) 0

Solution :-

Here

We know that

When a + b + c = 0, we get

Here,

So using the above identity, we get:

Now

{from equation i}

(ii) 0

Solution :-

We know that

When a + b + c = 0, we get

Here,

So using the above identity, we get:

....................(i)

Now

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
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• Faculty Support #### Without finding the cubes, factorize

solution :

given

We know that

When a + b + c = 0, we get

Here,

So using the above identity, we get:

#### Without actually calculating the cubes, find the value of(i) (ii)

(i)

Given

We know that

When a + b + c = 0, we get

Here,

So using the above identity, we get:

(ii)-0.018

Solution

Given:

We know that

When a + b + c = 0, we get

Here,

So,

## Crack NEET with "AI Coach"

• HD Video Lectures
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• Faculty Support #### Factorize(i) (ii)

(i)

Solution:

Given:

This can be written as:

We know that

Using this identity we get

(i)

Solution:

Given:

This can be written as:

We know that

Using this identity we get

=

#### Find the following product

Solution

Given:

This can be written as

We know that

Comparing equation (i) with the above identity we get:

a = 2x, b = -y, c = 3z

Hence,

## Crack JEE Main with "AI Coach"

• HD Video Lectures
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• Faculty Support #### Factorise:(i) (ii)

(i)

Given

we know that

so

(ii)

Given

This can be written as

we know that

so

(i)

Given:

(ii)

Given

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
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• Faculty Support #### Factorize the following :

(i)(1−4a)(1−4a)(1−4a)

Given, 1−64a3−12a+48a2

The given equation can be written as:

=(1)3−(4a)3−3(1)2(4a)+3(1)(4a)2           ..........(i)

Now we know that:

a3−b3−3a2b+3ab2=(a−b)3

Comparing equation (i) with the above identity, we get:

∴(1)3−(4a)3−3(1)2(4a)+3(1)(4a)2

=(1−4a)3

Hence the factorized form is (1−4a)(1−4a)(1−4a)

(ii)

Given,

The given equation can be written as:

…(i)

Now we know that:

Comparing equation (i) with the above identity, we get:

Hence the factorized form is

#### Expand the following :

Solution: Given

We know that

So we get:

Hence the expanded form is

Solution: Given

We know that

So we get:

Hence the expanded form is

Solution: Given

We know that

So we get:

Hence the expanded form is  .  