show that 1,2 are the zero's of the polynomial y^3-6y^2+11y-6
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Step-by-step explanation:
f(y) = y³-6y²+11y-6
f(1) = 1³-6(1)²+11(1) - 6
f(1)=1-6+11-6
f(1) = 12-12
f(1) = 0
By factor theorem 1 is a zero of polynomial.
f(2) = 2³-6(2)² + 11(2) - 6
f(2) = 8 - 24 + 22 - 6
f(2) = 30 - 30
f(2) = 0
By factor theorem 2 is a zero of polynomial.
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