Q. 2 P

# Let’s resolve into factors

x^{2} + 4abx-(a^{2}-b^{2})^{2}

Answer :

x^{2} + 4abx-(a^{2}-b^{2})^{2} … (1)

(a^{2}-b^{2})^{2}=[(a + b)^{2}(a-b)^{2}]^{2}

Given, x^{2} + (p + q)x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q = 4ab

pq =-(a^{2}-b^{2})^{2}

we know that (a + b)^{2}=a^{2} + 2ab + b^{2} … (3)

similarly, (a-b)^{2}=a^{2}-2ab + b^{2} … (4)

subtracting 4 from 3

(a + b)^{2}-(a-b)^{2} =a^{2} + 2ab + b^{2}-( a^{2}-2ab + b^{2})

=4ab

∴ (1) can be written as

x^{2} + (a + b)^{2}x-(a-b)^{2}x-[(a + b)^{2}(a-b)^{2}]^{2}

=[{x + (a + b)^{2}}{x-(a-b)^{2}}]

Apply the formula (a+b)^{2} = a^{2} + b^{2} + 2ab

(a-b)^{2} = a^{2} + b^{2} - 2ab

=[{x + a^{2} + b^{2} + 2ab }{x-( a^{2} + b^{2} - 2ab )}]

=[{x + a^{2} + b^{2} + 2ab }{x- a^{2} - b^{2} + 2ab }]

∴ the resolved factors are [{x + a^{2} + b^{2} + 2ab }{x- a^{2} - b^{2} + 2ab }]

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