Factorise:
$4 y^{2} + 20 y + 25$

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Solution

$4 y^{2} + 20 y + 25 = (2 y)^{2} + 2 \times 2 y \times 5 + (5)^{2} = (2 y + 5)^{2} {∵ a^{2} + 2 a b + b^{2} = (a + b)^{2}}$

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Similar questions

Evaluate: $(2 y + 5) (2 y + 5)$

Using an identity expand :
$(2 y + 5) (2 y + 5)$

(2 y + 5) (2 y + 5)

20 (y + 4) (y^{2} + 5 y + 3) \div 5 (y + 4)

20 (y + 4) (y^{2} + 5 y + 3) \div 5 (y + 4)

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