The graph of f(x) = ax^4 + bx^2 + cx-24 crosses the x-axis at 1, -2, and 3. determine the equation of f(x).

Question

anonymous · Answer

start with 
$$(x-1)(x+2)(x-3)(x-r)$$ and see if you can solve for $r$

anonymous · Answer

do i expand and then solve?

anonymous · Answer

no you don't really have to expand yet, you can visualize what the constant should be 
the constant will be $-6r$ so you know $-6r=-24$ and if you are lucky $r=4$ works

anonymous · Answer

oh damn it doesn't because you get a cube term and your polynomial doesn't seem to have one

anonymous · Answer

on the other hand something is wrong, because you have a 4th degree polynomial. it has 4 zeros total, although either two or four could be complex.  however, if it crosses the $x$ axis 3 times it must cross it a fourth.
if one root is repeated, then it does not cross the axis but only touches it.  
so i am stumped

anonymous · Answer

maybe one passes through origin?

anonymous · Answer

then it still has 4 zeros one would be zero, and also it could not end in $-24$

anonymous · Answer

it really starts with $ax^4$ right?

anonymous · Answer

yeah

anonymous · Answer

i have the answer if that helps?

anonymous · Answer

ok i am very certain that a fourth degree polynomial cannot cross the $x$ axis only 3 times

anonymous · Answer

sure go ahead and let me see why i am wrong

anonymous · Answer

its f(x)= -4x^3 + 8x^2 + 20 x -20

anonymous · Answer

we could perhaps start with 
$$a(x-1)(x+2)(x-3)(x-r)$$ and see if we can solve for the coefficients, but it must cross 4 times

anonymous · Answer

f(x)= -4x^3 + 8x^2 + 20 x -24 *** sorrry

anonymous · Answer

hold the phone, it is cubic?

anonymous · Answer

the answers a cubic :S

anonymous · Answer

i thought it was a 4th degree polynomial, since you wrote $$f(x) = ax^4 + bx^2 + cx-24$$

anonymous · Answer

if it is 
$$f(x) = ax^3 + bx^2 + cx-24$$ then that is a whole different story

anonymous · Answer

lol thats what the question says, its ax^4 + bx^2 ..

anonymous · Answer

and much simpler 
write 
$$a(x-1)(x-3)(x+2)= ax^3 + bx^2 + cx-24$$ and solve

anonymous · Answer

then there is something wrong with the question, it is a typo

anonymous · Answer

so if its ax^3 will the answer be right?

anonymous · Answer

your question asks for a 4th degree polynomial, which must have 4 zeros. it also has no cube term
your answer has a third degree polynomial, so there is a mistake in the question

anonymous · Answer

yeah it prob is a typo.. ah been trying to question for so long.. thanks tho!