let g(x) = x + 2a and p(x) = x⁵ – 4a²x³ + 2x + 2a + 3

Putting g(x) = 0

⟹ x + 2a = 0 ⟹ x = – 2a

According to the factor theorem if g(x) is a factor of p(x) then p( – 2a) = 0

Now p ( – 2a) = ( – 2a)⁵ – 4a²( – 2a)³ + 2( – 2a) + 2a + 3 = 0

⟹ – 32a⁵ + 32a⁵ – 2a + 3 = 0 ⟹ – 2a = – 3

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