Question

Find a third–degree polynomial function such that f(0) = –12 and whose zeros are 1, 2, and 3. Using complete sentences, explain how you found it.

Answer 1

it's gonna be of the form: c(x-1)(x-2)(x-3) for some constant c (why?)
and what do you think c should be given the rest of the information (f(0=-12))?

Answer 2

wait what?

Answer 3

so a zero of a function means a number N such that f(N) = 0.
f is a polynomial, so you can factor it. f(N)=0 implies that one of the factors is x-N.
there are three zeroes mentioned in the question: 1, 2, and 3. So f(x) can be factored into something times (x-1)*(x-2)*(x-3).

Answer 4

so do i just multiply them? or what?

Answer 5

well you're not quite done yet. you still need to satisfy the condition f(0) = -12.
if f(x) were simply (x-1)(x-2)(x-3) then f(0) would not be -12

Answer 6

then what do i do with the f(0)=-12?

Answer 7

well i haven't given you the final answer. how would you modify it so that f(0)=-12?

Answer 8

i really dont know...would you multiply it all by -12? im just guessing

Answer 9

why don't you try it out and see if it works?

Answer 10

okay i did that but i dont know

Answer 11

ugh... what did you get when you multiply it by -12?

Answer 12

-12x^3+72x^2-132x+72

Answer 13

well, now you need to check if it satisfies all the conditions given in the problem.
but out of curiosity, why did you expand the entire expression? (would it make a difference if you just used f(x) = -12(x-1)(x-2)(x-3) for now and expand it later after you checked that it works?)

Answer 14

i dont know...the reason i put it on here is because i dont know what to do..obviously?

Answer 15

my point was that you didn't need to expand it. it's the same function whether you call it (x-1)(x-2)(x-3) or x^3-6x^2+11x-6.

Answer 16

which way of expressing it would make your life easier when you're checking that it satisfies all the conditions?

Answer 17

ok, you were very close with your guess of multiplying by -12.
so now your function looks like: f(x) = -12*(x-1)(x-2)(x-3).
the question is, does this f(x) work?
i.e. does it satisfy f(0) = -12, and does it have three zeroes at 1, 2, and 3?

Answer 18

i guess?

Answer 19

ugh... it's fairly easy to check... just plug and chug

Answer 20

start with the easy ones: does f(1) = 0?
does f(2) = 0?
does f(3) = 0?

Answer 21

im seriously confused..