Factorise : (a + 2b) (3a + b) - (a + b)(a + 2b) + (a + 2b)^2

Question

KnowledgeBoat · Accepted Answer

(a \+ 2b) (3a \+ b) \- (a \+ b)(a \+ 2b) \+ (a \+ 2b)² = (a \+ 2b)(3a \+ b) \- (a \+ b) \+ (a \+ 2b) = (a \+ 2b) 3a \+ b \- a \- b \+ a \+ 2b = (a \+ 2b) (3a \- a \+ a \+ b \+ 2b \- b) = (a \+ 2b) (3a \+ 2b) Hence, (a \+ 2b) (3a \+ b) \- (a \+ b)(a \+ 2b) \+ (a \+ 2b)² = (a \+ 2b) (3a \+ 2b)

Mathematics

Factorise :
(a + 2b) (3a + b) - (a + b)(a + 2b) + (a + 2b)²

Factorisation

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Mathematics

Factorise :(a + 2b) (3a + b) - (a + b)(a + 2b) + (a + 2b)2

Factorisation

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(a + 2b) (3a + b) - (a + b)(a + 2b) + (a + 2b)²

Factorise :
12abc - 6a²b²c² + 3a³b³c³

Factorise :
4x(3x - 2y) - 2y(3x - 2y)

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