I need some help with this problem. Here is the link. Can you please tell me if there is an easier way to show that cubic polynomials have a real root? The question is in an analysis book from the continuity section so it has to use that. Here is the latex:

Show that a cubic equation (i.e. one of the form $$ax^3 + bx^2 + cx + d = 0$$ where $$a\neq 0)$$ has at least one real root.

Solution: The equation has at least one root if for some $$x_1, $$\enspace f(x_1) < 0$$ and $$f(x_2) > 0$$. Then by the intermediate value theorem $$f(c) = 0$$ for some $$x_1 < c < x_2$$. $$x^3$$ outgrows smaller powers of $$x$$ so the function is negative for some large negative number and positive for some large positive number. If $$(x_n)$$ is a sequence of positive terms that tends to infinity, then $$f(x_n) = ax_n^3 + bx_n^2 + cx_n + d = x_n^3(a+ \frac{b}{x_n} + \frac{c}{x_n^2} + \frac{d}{x_n^3})$$ Now $$\frac{b}{x_n}, \frac{c}{x_n^2}, \frac{d}{x_n^3}$$ are sequences that tend to zero, so for any $$\epsilon$$ there is an $$N$$ such that $$|\frac{b}{x_n}| < \epsilon/3, \quad |\frac{c}{x_n^2}| < \epsilon/3, \quad |\frac{d}{x_n^3}| < \epsilon/3$$ and for $$\epsilon = a$$, we have $$|\frac{b}{x_n}| + |\frac{c}{x_n^2}| + |\frac{d}{x_n^3}| < a$$ so that, by the triangle inequality $$|\frac{b}{x_n} + \frac{c}{x_n^2} + \frac{d}{x_n^3}| \leq |\frac{b}{x_n}| + |\frac{c}{x_n^2}| + |\frac{d}{x_n^3}| < a$$ which means $$-a <\frac{b}{x_n} + \frac{c}{x_n^2} + \frac{d}{x_n^3} < a$$ Then for some $$|k|<1$$, it can be written $$a+ \frac{b}{x_n} + \frac{c}{x_n^2} + \frac{d}{x_n^3} = a+ ka = (1+k)a$$ and $$f(x_n) = x_n^3(1+k)a$$ for $$n\geq N$$. Since $$x_n$$ is a sequence of positive terms, $$f(x_n) = k_na$$ for $$n\geq N$$ where $$k_n>0$$. If $$x_n$$ is instead chosen as a sequence of negative terms that tends to $$-\infty$$, then $$f(x_n) = (k_n')a$$ for $$n\geq N$$ where $$k_n'<0$$. Therefore regardless of the sign of $$a$$ the function $$f$$ takes on both positive and negative values.

It seems redundant and too many steps. Is there a more simple way to solve this problem? Any feedback is appreciated. Thank you!

Assume wlog $$a=1$$ by factoring $$a$$ out, which is doable since $$a\ne0$$.

Simpler inequalities can be deduced by taking $$x\ge|b|+|c|+|d|+1$$ so that we have

$$x+b,x+c,x+d\ge1$$

and noticing that we then have

\begin{align}f(x)&=x^3+bx^2+cx+d\\&=(x+b)x^2+cx+d\\&\ge(x+c)x+d\\&\ge x+d\\&\ge1\end{align}

and similarly that if we have $$x\le-(|b|+|c|+|d|+1)$$ then

\begin{align}x+b,x+c,x+d&\le-1\\-x+b,-x+c,-x+d&\ge1\end{align}

which then gives

\begin{align}f(x)&=x^3+bx^2+cx+d\\&=(x+b)x^2+cx+d\\&\le(-x+c)x+d\\&\le x+d\\&\le-1\end{align}

Q.E.D.

I agree, the proof is needlessly complex. These parts of it are almost enough:

The equation has at least one root if for some $$x_1, $$\enspace f(x_1) < 0$$ and $$f(x_2) > 0$$. ... $$x^3$$ outgrows smaller powers of $$x$$ so the function is negative for some large negative number and positive for some large positive number.

I say "almost", because the last sentence is not quite true. The function is negative for some large negative number if $$a$$ is positive. But the cubic polynomial $$-2x^3$$ (where $$a = -2$$) is positive for all negative numbers and negative for all positive numbers.

But if you factor $$a$$ out of the equation you are left with a monic polynomial that is negative for some large negative number and positive for some large positive number. So let $$x_1$$ be a large negative number that makes the polynomial negative, and let $$x_2$$ be a large positive number that makes the polynomial positive.

If you need to show how to explicitly exhibit values of $$x_1$$ and $$x_2$$ for any given values of $$a, b, c, d,$$ you already have another answer that shows how to do that.