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?
Best Answer
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.