why $x^3\space +\space ax^2\space -x\space -2$ can only have one solution greater than zero

Question

what is the proof for the above equation always having exactly one solution greater than zero for all values of a

I cannot see how to prove this because you cannot factorise the polynomial and I am not sure whether looking at its turning points will help

all I can see is that it crosses the y axis at y=-2

Answer 1

Suppose $f$ has roots $c_1$, $c_2$, and $c_3$. Then we can write:

$$f(x) = (x-c_1)(x-c_2)(x-c_3)$$

Expanding this out, we get:

$$f(x) = x^3 - (c_1+c_2+c_3)x^2 + (c_1c_2+c_1c_3+c_2c_3)x - c_1c_2c_3$$

Notice that the constant term is the signed product of the roots, which is negative in your problem.

Further, the coefficient on $x$ is negative in your problem.

Combining these facts, what can you deduce about the number of positive roots?

Answer 2

Note that if $f(x)=x^3+ax^2-x-1$ you have $f(0)=-2$ and $f'(0)=-1$.

Now for $x$ with large absolute value $f'(x)\approx 3x^2$ is positive. So $f'$ changes sign at least once for negative $x$ and at least once for positive $x$ - and since it has at most two changes of sign, it changes precisely once for positive $x$. This must be at a point for which $f(x)$ is negative.

You should be able to conclude from there.