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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?

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

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?

yeah

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

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?

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