Question

factorise a2-(2b+3c)2.and (4a+7b)2-(3a+8b)2.and (a+b+c)2-(a+b-c) 2

Answer 1

An Important formula is to be used in every question
x²-y²=(x+y)×(x-y)

Factorizing a²-(2b+3c)²

=> (a+2b+3c)×(a-2b-3c)
Factorized

Factorizing (4a+7b)² - (3a + 8b)²
=> (4a +7b+3a + 8b)×(4a+7b-3a-8b)
=> (7a+15b)×(a-b)

Factorizing (a+b+c)²-(a+b-c)²
=> (a+b+c+a+b-c)×(a+b+c-a-b+c)
=> 4×(a+b)×c

Answer 2

HELLO FRIEND

(1) a^2 - (2b + 3c )^2

As it is of the form a^2 - b^2 which is equal to (a+b) (a- b)

Then

Above question can be written as

{ a + (2b + 3c)} { a- (2b + 3c)}

(a + 2b + 3c) ( a - 2b -3c)


(2) (4a + 7b)^2 - (3a + 8b )^2


It is also of the type a^2 -b^2

Then,

{ (4a +7b) + (3a + 8b) } { (4a+7b) - (3a + 8b) }

(4a +7b + 3a + 8b) (4a +7b -3a - 8b)

(7a + 15b) (3a - b)


(3) (a + b + c )^2 - ( a + b - c) ^2

It is also of the type a^2 - b^2

Hence

{ (a + b + c ) + ( a + b - c) } { ( a + b + c ) - ( a + b - c) }

( a + b + c + a + b - c) ( a + b + c - a - b + c)

(2a + 2 b) (2c )

2 (a + b )(2c )

4 (a + b ) c


I HOPE THIS MAY HELP U