Question

ab^2-ba^2-ab+a^2 - Factorise

Answer 1

[tex] ab^{2} - ba^{2}-ab+ a^{2} =0 \\ Taking \ a \ common, \\ a( b^{2}- ab-b+a)=0 \\ a ( b^{2} -b-ab+a)=0 \\ a(b(b-1)-a(b-1))=0 \\ (a)(b-a)(b-1)=0[/tex]

Hence, (a), (b - a) and (b - 1) are the factors.

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