Question

Can you guys tell me if i have this right?

polynomial Function f(X) has given zero. Find the other zero

F(x) = x^3+2x^2-6x-12; -2

The other zeros are : -2,-(1/2),(1/3)

Answer 1

Well, if you plug the prospective zeros into f(x), you should get 0 as the result. Do you?

\[(\frac{1}{3})^3+2(\frac{1}{3})^2-6(\frac{1}{3})-12 = \frac{1}{27}+\frac{2}{9}-2-12\]I don't think that's going to simplify to 0, do you?

Answer 2

So at least (1/3) is not a zero. Also, you'll have exactly 3 zeros, because x^3 is the highest exponent you have. Your answer suggests that you think the zeros include -2 with multiplicity 2, but that's not correct, as that would give you 4 zeros instead of 3.

As you know 1 zero to start, you can simplify the polynomial by dividing it by (x-zero) to get a polynomial which has the same remaining roots but is lower in degree. So here you would divide by \((x-(-2)) = (x+2)\) and see what you get...

Answer 3

You might be able to see how to factor the remaining polynomial, or guess at a zero (do you know the rational root theorem?)

Answer 4

:*, i have no idea!!

Answer 5

First things first: what do you get when you divide \[x^3+2x^2-6x-12\]by\[x+2\]?