What is the alternating sum of coefficients and what does it have to do with the zeroes of the function?

Question

So, my teacher told us today, while we were solving this integral: $$\int\frac{dx}{x(2x^3+x^2+1)}$$ that the alternating sum of coefficients of $2x^3+x^2+1$ is 0 (2-1+0-1=0) and hence, one zero of the function is $-1$. Why is the zero $-1$?

Answer 1

This method can be used to check whether $-1$ is a root. If you plug $-1$ into a polynomial, you are actually adding the coefficients with alternating signs. That is because odd power of $-1$ is $-1$ and even power of $-1$ is $1$. So if you add alternating coefficients and get $0$, it means $-1$ is a root.

Answer 2

$$2x^3+x^2+1=(x+1)(2x^2-x+1)$$ and $2x^2-x+1\neq 0$ for all $x\in \mathbb R$. Therefore, the only zero of $x(2x^3+x^2+1)$ are $0$ and $-1$.

Answer 3

Here is a very non-rigorous way of developing the intuition behind Descartes' Rule of Signs.

Let $P(x)=2x^3+x^2+1$. For $x\ge 0$, $P(x)>0$.

For values of $x<<-1$, the leading term $2x^3$ dominates the quadratic and constant terms. $P$ decreases without bound and we must have a point $x_N$ for which $P(x)<0$ for all $x<x_N$ .

Since $P$ always increases, is greater than zero when $x>0$, is less than zero when $x<<-1$, then there must be one and only one point $x_0 <0$ for which $P(x_0)=0$.