show that 1/3 and 4/3 are zeroes of polynomial 9x^3-6x^2-11x+4 find third 0 of polynomial
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Step-by-step explanation:
p(1/3)=9(1/3)³-6(1/3)²-11(1/3)+4
=9(1³/3³)-6(1²/3²)-11(1/3)+4
=9(1/27)-6(1/9)-11(1/3)+4
=(1/3)-2(1/3)-11(1/3)+4
=(1/3)-(7/3)-(34/3)+4
=[(1-7-34)/3]+4
=(-40/3)+4
=[(-40+4)/4
=36/4
=9
Sorry I can't show that 1/3 is zeroes of the polynomial
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