Question

(ac+bd)^2-(ad + bc)^2​

Answer 1

Question

Factorise :

(ac+bd)²-(ad+bc)²

Solution

we have

→ (ac+bd)² - (ad+bc)²

using identities

( x + y )²= x² + y² + 2xy

→ (ac)²+(bd)²+2abcd - ( (ad)²+(bc)²+2abcd )

→ (ac)²+(bd)²+2abcd- (ad)²-(bc)²-2abcd

→ a²c² + b²d² - a²d² - b²c²

→ a²( c² - d² ) - b² ( c² - d² )

taking c²-d² common

→ ( c² - d² ) ( a² - b² )

using identity

x² - y² = ( x + y ) ( x - y )

→ ( c + d ) ( c - d ) ( a + b ) ( a - b )

Factorised .

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