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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 $x_1<x_2$, $\enspace f(x_1) < 0$ 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!

Simply Beautiful Art
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marpo
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2 Answers2

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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.

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

The equation has at least one root if for some $x_1<x_2$, $\enspace f(x_1) < 0$ 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.

David K
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