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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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start with \[(x-1)(x+2)(x-3)(x-r)\] and see if you can solve for \(r\)
do i expand and then solve?
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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oh damn it doesn't because you get a cube term and your polynomial doesn't seem to have one
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
maybe one passes through origin?
then it still has 4 zeros one would be zero, and also it could not end in \(-24\)
it really starts with \(ax^4\) right?
i have the answer if that helps?
ok i am very certain that a fourth degree polynomial cannot cross the \(x\) axis only 3 times
sure go ahead and let me see why i am wrong
its f(x)= -4x^3 + 8x^2 + 20 x -20
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
f(x)= -4x^3 + 8x^2 + 20 x -24 *** sorrry
hold the phone, it is cubic?
the answers a cubic :S
i thought it was a 4th degree polynomial, since you wrote \[f(x) = ax^4 + bx^2 + cx-24\]
if it is \[f(x) = ax^3 + bx^2 + cx-24\] then that is a whole different story
lol thats what the question says, its ax^4 + bx^2 ..
and much simpler write \[a(x-1)(x-3)(x+2)= ax^3 + bx^2 + cx-24\] and solve
then there is something wrong with the question, it is a typo
so if its ax^3 will the answer be right?
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
yeah it prob is a typo.. ah been trying to question for so long.. thanks tho!

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