higher order polynomials

Question

Suppose all the roots of the equation $x^3+bx-2017=0$ (where $b \in \mathbb{R}$) are real. Prove that exactly one root is positive. My progress:

• As product of three roots is $2017$ which is positive integer. (then possibilities are either any two are negative and 1 is positive or all three are positive.)
• And the sum of three roots is $0$. (which means either all of them are zero or few are negative and positive later one is not possible as $0$ is not solution to above problem and product of roots is $2017$)
• I also found out that $b<0$.
  From above three we can say that one root has to be positive. Am I correct?

Is there any other way of doing above problem? I am sure I can find one here.

Thank you

Answer 1

Your answer is correct. The product of the roots is positive, hence at least one is positive (because there are three of them). Their sum is zero, so this implies that at least one root is negative. Again, since the product is positive, the only possibility is that one is positive and the other two are negative.

Answer 2

Look at the graph of $f(x)=x^3+bx-2017$.

Let the roots of $f$ be $r_1 < r_2 < r_3$.

Since $f(0)<0$, we have $0 \in (-\infty,r_1) \cup (r_2,r_3)$.

Since $r_1 + r_2 + r_3 = 0$, we have $r_1 < 0$ and so $0 \in (r_2,r_3)$.

This means that $r_1 < r_2 < 0 < r_3$ and so $r_3$ is the only positive root.

Answer 3

Your approach is fine. Here is another one:

By Descartes’ rule of signs, there is exactly one sign change in the polynomial (irrespective of what $b$ is), so it has exactly one positive root.

Notice we didn’t have to use the fact that all roots are real, all we needed was $b\in \mathbb R$.