# Algebrically determining when a cubic function passes through certain points and has a value of $120$

Algebraically find where the cubic polynomial function that has zeroes at $2, 3 -5$ and passes through $(4, 36)$, has a value of $120$.

Yeah, so this is a question in my textbook which I don't really understand what its asking. It's an inequality question. The textbook answer is $x =-2, x =-3, x =5$.

I try writing up the equation as $f(x) = (x-2)(x-3)(x+5)$. Then I am not too sure where to go from there since I have $2$ $y$-values. So can anyone tell me how to solve this functions inequality?

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What do you mean by "pass through 4, 36 has a value of 120"? How is "-2, -3, -5" even a polynomial at all? Also, your f(x) has a zero at +5 instead of -5. – anon Oct 12 '11 at 5:16
yeah i think i should of said (4,36) instead of "4,36" as for a value of 120 i am pretty sure thats the y value it dosent say in the question though. – Faraz Oct 12 '11 at 5:21

You want to start with $(x-2)(x-3)(x+5)$ to make sure the zeroes are correct. Your extra degree of freedom comes from the fact that $c(x-2)(x-3)(x+5)$ also has those roots for any constant $c$.

If the function is to pass through $(4,36)$, then you want to solve $$c(4-2)(4-3)(4+5) = 36$$ for $c$. I get $c = 2$.

Your function is entirely specified now: $f(x) = 2(x-2)(x-3)(x+5)$.

To find out where it has a value of 120, you want to solve $$120 = 2(x-2)(x-3)(x+5)$$ for $x$. Depending on what options you have at your disposal, you can do this either by graphing or by setting the function equal to 0 and using a combination of rational roots tests, polynomial division, and factoring.

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yeah i got the answer except one minor difference the c value would be 2 since 36 = a(4-2)(4-1)(4+5) 36 =18a a =2 for the rest of the roots i just solved for when both sides =0 after moving 120/2 over then syntehtic divison (a, c i assume they both mean the same) – Faraz Oct 12 '11 at 5:38
I get $c = 2$ as well. I flubbed up some numbers on my first try. – Austin Mohr Oct 12 '11 at 5:41

After Austin Mohr's answer, expand the last equation $$2(x-2)(x-3)(x+5)-120 = 2 x^3-38 x-60$$

By inspection $x=-2$ is a root. So, perform the long division and you will have a quadratic equation.

I am sure you can take from here.

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