I was asked to prove that any cubic polynomial with real coefficients has at least one real root. I am not very good at proving things but I made an attempt, It was a bit of a draft and not perfect also I didn't know how else to say that if the cubic term is larger than all the other terms combined then that value for x would hold regardless of the values for the other terms, so if someone could help me with this I'd really appreciate it, or if someone could help me improve it, tell me how I can fix some of my errors, or show me a more elegant way to prove it. Thanks! :)
Homework proof of cubic polynomial
411 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtThere are 2 best solutions below
On
A more elegant way to prove this may be the following:
Since the polynomial has real coefficients, then, by the complex conjugate root theorem, if there is a non-real root, it's conjugate will be a root as well. Thus, there has to be an even number of non-real roots. Since the polynomial has 3 roots, there can be at most two non-real roots. It follows that there is at least one real root. Otherwise, there will be 3 real roots.
If you use the intermediate value theorem,
For $f(x)=ax^3+bx^2+cx+d=0$, assume $a>0$. This can be done without loss of generality, because if $a<0$, both sides can be multiplied by $-1$ to make the coefficient of $x^3$ positive.
Now, you can apply IVT: $\lim_{x \rightarrow +\infty}f(x)=+\infty$ and $\lim_{x \rightarrow -\infty}f(x)=-\infty$. Since $f$ is continuous everywhere, there is at least one root in the range $(-\infty,+\infty)$


This comes directly as a consequence of the Intermediate Value Theorem, which states that if $a,b\in\mathbb{R}$ such that $a<b$, and the function f is continuous on the closed interval [a,b], then $\forall{y_0}$ where $f(a)<y_0<f(b)$ ,$\exists{x_0}\in[a,b]$ with $f(x_0)=y_0$. So if you choose an extremely large negative value for $a$, an extremely large value for $b$ and set $y_0=0$, which follows $f(a)<0<f(b)$, $\exists{x_0}\in[a,b]$ with $f(x_0)=0$. Since $x_0\in[a,b]$, we can also say $x_0\in\mathbb{R}$, completing the proof.