Question

factorize 4a^2 - 9c^2 - 25 - 12ac

Answer 1

The question will be :- Factorize 4a² + 9c² - 25 - 12ac

 4a²+9c²-25-12ac

= (2a)²+(3c)²-(5)²-2.2a.3c

Here we will use { (a -b)²( a² +2ab - b²)}

= [(2a)²+2.2a.3c- (3c)]²-(5)²

={(2a-3c)²} - (5)²

next  step we will follow:- ( a² - b²) = ( a +b ) ( a - b)

=(2a-3c+5) (2a-3c-5)

Ans :- (2a-3c+5) (2a-3c-5)

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Answer 2

$\displaystyle \sf{4 {a}^{2} + 9 {c}^{2} - 25 - 12ac} = (2a - 3c + 5)(2a - 3c - 5)$

Correct question : Factorise 4a² + 9c² - 25 - 12ac

Given :

$\displaystyle \sf{4 {a}^{2} + 9 {c}^{2} - 25 - 12ac}$

To find :

To factorise the expression

Identity Used :

$\displaystyle \sf{ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab}$

$\displaystyle \sf{ {a}^{2} - {b}^{2} = (a + b)(a - b) }$

Solution :

Step 1 of 2 :

Write down the given expression

Here the given expression is

$\displaystyle \sf{4 {a}^{2} + 9 {c}^{2} - 25 - 12ac}$

Step 2 of 2 :

Factorise the expression

$\displaystyle \sf{4 {a}^{2} + 9 {c}^{2} - 25 - 12ac}$

$\displaystyle \sf{ = {(2a)}^{2} + {(3c)}^{2} - 12ac - 25}$

$\displaystyle \sf{ = {(2a)}^{2} + {(3c)}^{2} - 2.2a.3c - 25}$

$\displaystyle \sf{ = {(2a - 3c)}^{2} - 25} \: \: \: \bigg[ \: \because \:{(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \bigg]$

$\displaystyle \sf{ = {(2a - 3c)}^{2} - {(5)}^{2} }$

$\displaystyle \sf{ = (2a - 3c + 5)(2a - 3c - 5)}\: \: \: \bigg[ \: \because \: {a}^{2} - {b}^{2} = (a + b)(a - b) \bigg]$

$\boxed {\: \:\displaystyle \sf{ \therefore \: \: 4 {a}^{2} + 9 {c}^{2} - 25 - 12ac} = (2a - 3c + 5)(2a - 3c - 5) \: \:}$

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$1.$ Find factorisation x⁴+2x²+9

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