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What is the third–degree polynomial function such that f(0) = –18 and whose zeros are 1, 2, and 3?
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Ahh thank you!!!
wait, lol. I didn't see the f(0) = -18 part
oh. so what do i do im soooo confused!
you know it is \[f(x)=a(x-1)(x-2)(x-3)\] and you also know the constant is 18 since \[f(0)=18\] the only thing you don't know is "a"
\[a\times -1\times -2\times -3=-6a=18\] therefore \[a=-3\]
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and so your answer is \[f(x)=-3(x-1)(x-2)(x-3)=-3 x^3+18 x^2-33 x+18\]
@luis unfortunately when you tack on -18 at the end of the product, the zeros are no longer 1,2 and 3
satellite is the master
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lol i hate algebra!!!! its mean
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