Question

How would you find all the zeros of f(x)=3x^3-5x^2+48x-80?

Answer 1

do you have a graphing calculator?

Answer 2

Being a cubic, it has at least one real root. Note that it's negative at 0.

since the coefficient of x^3 is positive, the polynomial muxt be positive for large x. So it has a positive root.

p(1) = -34
p(2) = 20

so it's zero between 1 and 2

now the sum of the roots is -b/a = 5/3.

product of the roots = -d/a = 80/3.

Just as a guess, what if there was only one real root?
The sum of the roots is between 1 and 2 ... what if that was the root?

I.e. substitute 5/3 in.

Yes, you get 0. So the other roots sum to 0. And multiply to give 16.

You could figure it out from there, or you could do polynomial long division to remove the factor of (3x-5)

Or you can figure it out faster:

Note that the cubic = (3x-5) times a quadratic.

See that the x^2 coefficient in the quadratic is 1, and the constant coefficient is +16.

So (3x-5) (x^2 + ? + 16).

But we know that its roots sum to 0 which means the middle coefficient must be 0.

So the cubic must factor to (3x-5) (x^2 + 16).

(check by multiplying that back out).

Hence write down all three zeros.