Q. 2 P

# Let’s resolve int

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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