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Step-by-step explanation:
given 3,-1 and -1/3
are the zeroes of the cubic polynomial
p(3)=3x^3-5x^2-11x-3
=3(3)^3-5(3)^2-11(3)-3
=81-45-33-3
=0
p(-1)=3(-1)^3-5(-1)^2-11(-1)-3
=-3-5+11-3
=-8+11-3
=0
p(-1/3)=3(-1/3)^3-5(-1/3)^2-11(-1/3)-3
=3×-1/27-5/9+11/3-3
=-1/9-5/9+11/3-3
=0
therefore,the zeroes of the polynomial are 3,-1,-1/3
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