Factorisation of 3a^2-b^2-c^2+2ab-2bc+2ca
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3a²−b²−c²+2ab−2bc+2ac
=4a²−a²−b²−c²+2ab−2bc+2ac
=4a²−(a²+b²+c²−2ab+2bc−2ac)
=(2a)²−(a−b−c)²
=(2a+a−b−c)(2a−a+b+c)
=(3a−b−c)(a+b+c)
Is the answer
=4a²−a²−b²−c²+2ab−2bc+2ac
=4a²−(a²+b²+c²−2ab+2bc−2ac)
=(2a)²−(a−b−c)²
=(2a+a−b−c)(2a−a+b+c)
=(3a−b−c)(a+b+c)
Is the answer
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