Degree Polynomials and Zeroes

Question

"Find a degree $3$ polynomial that has zeros $-3, 4$ and $8$ and in which the coefficient of $x^2$ is $-18$."

I've been trying to solve this problem, but I keep getting it wrong. I've worked with other problems in which the highest degree of the polynomial is given a coefficient, and looked up examples for this one, but can't figure out a way to apply what I've learned to solve this problem.

Steps Done: \begin{align} f(x) &= (x+3)(x-4)(x-8)\\ f(x) &= (x^2-x-12)(x-8)\\ f(x) &= a(x^3-7x^2-4x+96) \end{align} Side Note
When I solved for "a", I ended up with "a = 2.57", but it says that that isn't the correct answer.

Answer 1

You're almost entirely correct.

Certainly, the polynomial will be some constant multiple of $f(x) = (x+3)(x-4)(x-8)$. It seems you've made one tiny error, however. Multiplying this out, the coefficient on $x^2$ should be $-9$, and not $-7$.

From there, you just solve $-9a = -18$, which unfortunately you were trying to do with the wrong coefficient.

Answer 2

The cubics that have the desired roots have the shape $k(x+3)(x-4)(x-8)$.

Note that the sum of the roots of $(x+3)(x-4)(x-8)$ is $9$, so the coefficient of $x^2$ is $-9$. We want it to be $-18$, so we must take $k=2$.