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What is the third–degree polynomial function such that f(0) = –18 and whose zeros are 1, 2, and 3? Answer f(x) = 2x3 – 12x2 + 18x – 18 f(x) = 3x3 + 18x2 – 33x – 18 f(x) = 2x3 + 12x2 – 18x – 18 f(x) = 3x3 – 18x2 + 33x – 18
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f(x) =3x3 – 18x2 + 33x – 18 =(x-1)(x-2)(3x-9) f(0)= -18 and its zeros are 1, 2, and 3
yup
f(x)=a(x-1)(x-2)(x-3) f(0)=-6a=-18, then a=3 Since f(x)=a(x-1)(x-2)(x-3)=f(x)=3(x-1)(x-2)(x-3) So f(x)=3x^3-18x^2+33x-18
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