x2 + 4abx-(a2-b2)2 … (1)(a2-b2)2=[(a + b)2(a-b)2]2Given, x2 + (p + q)x + pq=(x + p)(x + q) … (2)Comparing 1 and 2p + q = 4abpq =-(a2-b2)2we know that (a + b)2=a2 + 2ab + b2 … (3)similarly, (a-b)2=a2-2ab + b2 … (4)subtracting 4 from 3(a + b)2-(a-b)2 =a2 + 2ab + b2-( a2-2ab + b2)=4ab∴ (1) can be written asx2 + (a + b)2x-(a-b)2x-[(a + b)2(a-b)2]2=[{x + (a + b)2}{x-(a-b)2}]Apply the formula (a+b)2 = a2 + b2 + 2ab(a-b)2 = a2 + b2 - 2ab=[{x + a2 + b2 + 2ab }{x-( a2 + b2 - 2ab )}]=[{x + a2 + b2 + 2ab }{x- a2 - b2 + 2ab }]∴ the resolved factors are [{x + a2 + b2 + 2ab }{x- a2 - b2 + 2ab }]

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Class 8th West Bengal - Mathematics 13. Factorisation of Algebraic Expressions

Answer :

x² + 4abx-(a²-b²)² … (1)

(a²-b²)²=[(a + b)²(a-b)²]²

Given, x² + (p + q)x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q = 4ab

pq =-(a²-b²)²

we know that (a + b)²=a² + 2ab + b² … (3)

similarly, (a-b)²=a²-2ab + b² … (4)

subtracting 4 from 3

(a + b)²-(a-b)² =a² + 2ab + b²-( a²-2ab + b²)

=4ab

∴ (1) can be written as

x² + (a + b)²x-(a-b)²x-[(a + b)²(a-b)²]²

=[{x + (a + b)²}{x-(a-b)²}]

Apply the formula (a+b)² = a² + b² + 2ab

(a-b)² = a² + b² - 2ab

=[{x + a² + b² + 2ab }{x-( a² + b² - 2ab )}]

=[{x + a² + b² + 2ab }{x- a² - b² + 2ab }]

∴ the resolved factors are [{x + a² + b² + 2ab }{x- a² - b² + 2ab }]

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