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Show that the cubic eq:

$$x^3+ax^2+bx+c = 0 \quad a,b,c\in \mathbb{R}$$

has at least one real root.

I know that the above equation can be broken down into $(x-a)(x-b)(x-c) = 0$ , but I have no idea what to do next. I can't use IVT to do this because I don't have a specified range.

(edit): For others reading this, the equation CANNOT be broken down to $(x-a)(x-b)(x-c) = 0$

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  • $\begingroup$ The above equation certainly cannot be broken down into that form, even if the second set of $a,b,c$ is different from the first. As for solving the problem, look at its end behaviour at $\pm\infty$ and use the IVT on some sufficiently large interval. $\endgroup$ – Ian Coley Dec 15 '13 at 20:37
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    $\begingroup$ You don't need a specified range. Just choose $x$ large enough and small enough. $\endgroup$ – mjqxxxx Dec 15 '13 at 20:38
  • $\begingroup$ The way you broke up the equation means that it has roots of a,b and c. $\endgroup$ – Module Dec 15 '13 at 20:41
  • $\begingroup$ There needs to be a distinction: the polynomial certainly splits, as @Shuri says, over $\mathbb{C}$, but not necessarily $\mathbb{R}$. $\endgroup$ – Eric Thoma Dec 15 '13 at 20:55
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All odd polynomials have at least one real root because of the intermediate value theorem. To prove this just plug in a very large positive number and a very large negative number for $x$ (e.g. $10^{23}$ and $-10^{23})$ and note that corresponding $y$ values will have opposite signs. Then the IVT tells you that there is at least one value of $x$ between the two large numbers for which $y=0$ i.e., the polynomial has a root.

Another thing to note is that the only irreducible polynomials over the reals are quadratic and linear. Since this is a polynomial of degree three it has to be decomposible either as three linear factors or one irreducible quadratic factor with a linear factor.

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    $\begingroup$ The large numbers you require come from the $\lim_{n\to-\infty} f(n)=-\infty$, so there exists $N<0$ such that $f(n)<0$ for all $n<N$. $\endgroup$ – Ian Coley Dec 15 '13 at 20:43
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If all the roots are real, then it is proven. Otherwise, there is at least one non-real root $z$. But then $\bar{z}$, the conjugate, is also a root (can you show this?). Thus, there are an even number of non-real roots. Since there are three roots in $\mathbb{C}$, it follows that at least one must be real.

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Just take the first and second derivatives and solve them for f(x) = 0.

You will note that you will have either: a) different values for the maximum, minimum and inflection point b) identical values for the max, min and inflection point

For case A, you can interpret it as the function having two "changes of direction" (at the maximum point the function stops increasing and starts decreasing, and inversely for the minimum point). Even if both maximum and minimum points have the same sign, there has to be at least one intersection with the x-axis, either before or after the interval between the maximum and minimum points. If they have opposite signs, then there may be a root before, after and between the maximum and minimum interval.

For case B, all three points are the same so after that point the function will continue its original course so there must be one intersection anyway.

Alternatively (and maybe more rigurously) as EVERY -linear- cubic function (as the one stated in your question) can be represented as the product of a first and second order polynomial, non real roots can only be obtained from the second order polynomial. The root of the first order polynomial will always be real.

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Let $z=s+ti$, and $f(z)=0$.

Then $$z^3+az^2+bz+c=0$$ so that we have:

$$s^3+3s^2ti-3st^2-t^3i + a(s^2+2sti-t^2) + b(s+it) + c=0$$

Now consider $\overline z=s-it$.

$$f(\overline z)=s^3-3s^2ti-3st^2+t^3i + a(s^2-2sti-t^2) + b(s-it) + c$$

Only the sign of the imaginary component has changed, which equals $0$.

So if $z$ is a zero, so is $\overline z$.

As a polynomial has a number of zeroes equals to its degree, a cubic has at least one real root.

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