Discussion on answer by Adriano: Finding the equation of a cubic polynomial $f(x)$, then solving $f(x) = 120$.

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03:26

A: Finding the equation of a cubic polynomial $f(x)$, then solving $f(x) = 120$.

You should read it in two parts: Find the unique cubic polynomial $y = f(x)$ that has zeroes at $2$, $3$, and $-5$ and passes through the point $(4,36)$. Find all values of $x$ such that $f(x) = 120$. As a hint for the first part, note that the zeroes tell us that the function has t...

user190748

so, do I substitute 36 for "a" and evaluate it at f(4)? What of the 120 though, doesn't that not get substituted as the Y value?

Adriano

Substitute $x = 4$ and $y = 36$ in order to solve for $a$. Once you know what $a$ is, you can then substitute $y = 120$ in order to solve for $x$.

user190748

Oh okay, I understand. Thanks for the help!

To check if I have made an error, I got that x= 128-x2 which if you got the root of it would equal (x-11.31)(x+11.31) I think this might be incorrect though.

Adriano

Yes, you made an error. Did you get $a = 2$? As a hint, note that $f(5) = 120$. There are two other values of $x$ such that $f(x) = 120$.

user190748

Yes I got a=2. But how does that work? After finding that a=2 I was using 120=2(x-2)(x-3)(x+5) to try to isolate "x".

Adriano

03:26

Expand and bring everything to one side to get: $$ 0 = x^3 - 19x - 30 $$ then guess some factors (use the rational root theorem or factor theorem), then perform long division to factor completely.

user190748

Oh! I made an error when I expanded the brackets. I see it now

Okay I really don't know why my answer isn't getting x3-19x-30.

I got it to 120=2(x^3-9x+30) I know I still have to multiply the 2 through and subtract the 120, but I don't think that is the answer that will come up

if evaluated it comes to 0=2x^3 - 18x - 60

Adriano

Note that you should get a $-19x$ term (and not a $-9x$ term) because: $$ 6x - 10x - 15x = -19x $$ Once you have that, just divide both sides of the equation by $2$ and subtract the $60$ to the other side.

user190748

Sorry, I've been doing math all day and my brain is starting to "fry", I keep expanding the brackets and getting different answers. I just got 2(x^3-16x+20)

correction: +30*

could you show me your process to try and figure out what it is I am doing wrong?

Adriano

We have: \begin{align*} (x - 2)(x - 3)(x + 5) &= (x^2 - 5x + 6)(x + 5)\\ &= (x^3 - 5x^2 + 6x) + (5x^2 - 25x + 30)\\ &= x^3 - 19x + 30 \end{align*}

user190748

shouldn't it be (x^2+5x-2x=10)(x-3). I don't see how the -5x makes sense.

correction: (x^2+5x-2x-10)

wait never mind, the -3x-2x

Okay, I spotted my error. It was a simply calculation mix-up.

I found that it had a factor x+2 , however my remainder did not equal zero when I divided it by x^3-19x-30

sorry, haha I forgot to input the 0x^2, so never mind

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