Answer :

(i) Let p(x) =3π‘₯2 + 6π‘₯βˆ’24 and g(x) = x – 2


Zero of g(x) = 2


Here p(2) = 3(2)2 + 6 (2) – 24 = 12 + 12 – 24 = 0


So g(x) is a factor of p(x)


(ii) Let p(x) = 4π‘₯2 + π‘₯βˆ’2 and g(x) = x – 2


Zero of g(x) = 2


P(2) = 4(2)2 + 2βˆ’2 = 16 β‰  0


Hence g(x) is not a factor of p(x)


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