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Factorise :

a2+bcacb2a^2 + bc - ac - b^2

Factorisation

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Answer

a2+bcacb2=(a2b2)+(bcac)=(ab)(a+b)+(bcac)=(ab)(a+b)+c(ba)=(ba)(a+b)+c(ba)=(ba)[(a+b)+c]=(ba)[ab+c]=(ab)(a+bc)a^2 + bc - ac - b^2 = (a^2 - b^2) + (bc - ac)\\[1em] = (a - b)(a + b) + (bc - ac)\\[1em] = (a - b)(a + b) + c(b - a)\\[1em] = -(b - a)(a + b) + c(b - a)\\[1em] = (b - a)[-(a + b) + c]\\[1em] = (b - a)[- a - b + c]\\[1em] = (a - b)(a + b - c)

Hence, a2+bcacb2=(ab)(a+bc)a^2 + bc - ac - b^2 = (a - b)(a + b - c).

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