Question

What are the roots of x^3-6x^2+3x+10?

Answer 1

I'd try -1, -2, -5, and -10 in that order. Why?

Answer 2

how did you do that? I know you should factor this.. but I just know how to factor the quadratic.. not the polynomial. how?

Answer 3

There is a wy to factor ALL cubic polynomials. You should look it up some time, just for laughs. You do NOT want to use it regularly.

In this case, I first relied on Descartes Rule of signs.

If f(x) = x^3 - 6x^2 + 3x + 10,

We look at the signs on the coefficients of f(x)

+ - + +

How many times does it CHANGE. You should see that it is twice. This means there might be as many as TWO Positive Real zeros of f(x). There might also be zero (0) zeros, so that's not as encouraging as it could be.

Now, define f(-x) = -x^3 - 6x^2 - 3x + 10

We look at the signs on the coefficients of f(-x)

- - - +

How many times does it CHANGE. You should see that it is once. This is VERY encouraging as the MUST be a Real Negative Zero of f(x). We should start looking on the negative side, since there IS one there. There might not be on the positive side.

Let's see if you followed that far or if I'm speechifying for a different time and a different place.

Answer 4

And then? What should I do?

Answer 5

The next thing is really a leap of faith. Since it would be SO MUCH easier if a zero were Rational, let's hope there is a Rational zero! Rational Zeros are ALWAYS built like this:

\(\dfrac{Some\;Factor\;of\;the\;Constant\;Term}{Some\;Factor\;of\;the\;Leading\;Coefficient}\)

Since the leading coeffient is 1 (one), this leaves only INTEGERS - even better. The possible integer zeros are factors of 10, and since we're on the negative side of the Origin, we pick -1, -2, -5, -10. Those are the ONLY possibilities for negative Rational zeros.

In this case, we were VERY lucky as x = -1 was exactly what we wanted to find. Don't count on that happening every time!