What is the third–degree polynomial function such that f(0) = –18 and whose zeros are 1, 2, and 3?

Question

anonymous · Answer

Ahh thank you!!!

hero · Answer

wait, lol. I didn't see the f(0) = -18 part

anonymous · Answer

oh. so what do i do im soooo confused!

anonymous · Answer

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"

anonymous · Answer

$$a\times -1\times -2\times -3=-6a=18$$ therefore 
$$a=-3$$

anonymous · Answer

and so your answer is 
$$f(x)=-3(x-1)(x-2)(x-3)=-3 x^3+18 x^2-33 x+18$$

anonymous · Answer

@luis unfortunately when you tack on -18 at the end of the product, the zeros are no longer 1,2 and 3

hero · Answer

satellite is the master

anonymous · Answer

of disaster!

anonymous · Answer

lol i hate algebra!!!! its mean