Question

evaluate i) (3b-1)(3b+2)
ii) (2+a)(2-a)
iii) (4+5x)(4-5x)
iv) (x+1)(x-1) ​

Answer 1

$\sf{Answer}$

Evalute :-

• ( 3b - 1 ) ( 3b + 2 )
• (2 + a ) ( 2 - a)
• (4 + 5x ) ( 4 -5x )
• (x + 1 ) ( x - 1 )

Formula to know :-

( a + b) ( a - b ) = a² - b²

_________________________

1 ) ( 3b - 1 ) ( 3b + 2 )

We should not use the above formula because a, b are different

So, multiply as usual

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

3b ( 3b + 2 ) - 1 ( 3b + 2 )

9b² + 6b - 3b - 2

9b² + 3b - 2

_________________

2 ) ( 2 + a ) ( 2 - a)

It is in form o f( a + b) ( a - b ) = a² - b²

( 2 + a ) ( 2 - a) = (2)² - (a)²

(2 + a ) ( 2 - a ) = 4 - a²

4 - a²

___________________

3) ( 4 + 5x ) ( 4 - 5x)

It is in form of ( a + b) ( a - b ) = a² - b²

(4 + 5x ) ( 4 - 5x)

(4)² - (5x)²

16 - 25x²

______________________

4) ( x + 1 ) ( x -1 )

It is in form of ( a + b) ( a - b ) = a² - b²

(x + 1 ) ( x -1 )

(x)² - (1)²

x² - 1

________________________

Know more :-

( a + b)² = a² + 2ab + b²

( a - b)² = a² - 2ab + b²

a² + b² = ( a + b)² - 2ab

( a + b)² + ( a -b)² = 2a² + 2b²

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

If a+ b+ c = 0 then a³ + b³ + c³ =0

Answer 2

$\underline{\underline{\sf{\maltese\:\:To\: Evaluate}}}$

• (3b - 1)(3b + 2)
• (2 + a)(2 - a)
• (4 + 5x)(4 - 5x)
• (x + 1)(x - 1)

$\underline{\underline{\sf{\maltese\:Formula \:Used}}}$

• (a + b)(a - b) = a² - b²

$\underline{\underline{\sf{\maltese\:Calculations \:}}}$

$\sf\red{(3b-1)(3b+2)}$

It does not suit the identity [(a + b)(a - b) = a² - b²] because b is not same in the given equation, so we have to simply multiply it as usual.

By multiplying, we get :-

$\longrightarrow\sf \: 3b(3b + 2) \: \: - 1(3b + 2)$

$\longrightarrow\sf \: 9 {b}^{2} + 6b - 3b - 2$

$\longrightarrow \red{\sf \: 9 {b}^{2} + 3b - 2}$

_____________________________________________

$\sf \pink{(2+a)(2-a) }$

The above equation suits the identity [(a + b)(a - b) = a² - b²] as the value of a and b are same and the signs are also exactly matches the identity, so let's solve it !!.

Here,

a = 2 and b = a

Now, By using identity [(a + b)(a - b) = a² - b²],we get :-

$\longrightarrow\sf \:(2)^{2} - {(a)}^{2}$

$\longrightarrow\sf \pink{\:4 - {a}^{2} }$

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$\sf \purple{ (4+5x)(4-5x)}$

This equation also suits the identity which we using in the above equations, so let's solve this with that identity only.

Here,

a = 4 and b = 5

Now, by putting the identity [(a + b)(a - b) = a² - b²], we get :-

$\longrightarrow\sf {(4)}^{2} - {(5x)}^{2}$

$\longrightarrow\sf \purple{16 - 25 {x}^{2} }$

_____________________________________________

$\sf \orange{(x+1)(x-1)}$

This equation also suits the identity which is being used in the equations which we solved earlier, so let's solve this also with that.

Here,

a = x and b = 1

Now, by using the identity [(a + b)(a - b) = a² - b²],we get :-

$\longrightarrow\sf {(x)}^{2} - {(1)}^{2}$

$\longrightarrow\sf \orange{{x}^{2} - 1}$

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