No polynomial of degree 3 in $\mathbb{R}[x]$ is a prime

Question

How do I prove that no polynomial of degree 3 in the ring $\mathbb{R}[x]$ of polynomials with real coefficients is a prime?

I really need help on this one guys. I know I need to use the Intermediate Value Theorem but I'm not sure how to get started.

Answer 1

Let $f(x) = ax^3+bx^2+cx+d$ and assume, WLOG (without loss of generality), that $a>0$.

Then

$$\lim_{x\to-\infty}f(x)=-\infty$$

and

$$\lim_{x\to\infty} f(x) =+\infty$$

(WLOG since if $a<0$ then we still get one infinite and one negative infinite limit, which is all we need)

So $f(x)$ takes on positive and negative values, otherwise it couldn't approach $+\infty$ or $-\infty$ respectively, hence has a zero by the IVT since $0$ is between any positive and negative number. Since it is a polynomial, this means it factors as $(x-r)(a'x^2+b'x+c')$, i.e. it is reducible. This is seen because all real polynomials factor over $\Bbb C$, and since $f(x)=(x-r_1)(x-r_2)(x-r_3)$ for every complex $x$, the only way to have a product of three things be $0$ is if one of them is zero. Since this happens for some $x\in \Bbb R$ from the IVT, we determine the polynomial has a real root.

Answer 2

You have $f(x)=ax^3+\text{lower-degree terms}$. Show that if $x$ is large by comparison to $a$ and to the lower-degree coefficients, then $ax^3$ overwhelms the lower-degree terms, so that $f(x)\to+\infty$ as $x\to+\infty$ and $f(x)\to-\infty$ as $x\to-\infty$ (if $a>0$) and $f(x)\to-\infty$ as $x\to+\infty$ and $f(x)\to+\infty$ as $x\to-\infty$ (if $a<0$).

Since polynomial functions are continuous, if this function goes from $-\infty$ to $+\infty$ then someone in between it is equal to $0$. Suppose $f(k)=0$. Now remember something from algebra: if $f(k)=0$, then $f(x)$ is divisible by $x-k$, i.e. $f(x)=(x-k)(\cdots\cdots\cdots)$, where the dots represent some second-degree polynomial.