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- anonymous

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).

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- anonymous

- katieb

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- anonymous

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

- anonymous

do i expand and then solve?

- anonymous

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

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- anonymous

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

- anonymous

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

maybe one passes through origin?

- anonymous

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

- anonymous

it really starts with \(ax^4\) right?

- anonymous

yeah

- anonymous

i have the answer if that helps?

- anonymous

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

- anonymous

sure go ahead and let me see why i am wrong

- anonymous

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

- anonymous

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

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

- anonymous

hold the phone, it is cubic?

- anonymous

the answers a cubic :S

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

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

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