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While solving a problem a friend said - this polynomial is $3^{rd}$ order ($ax^3+bx^2+cx+d$), with $\{a,b,c,d\}$ real coefficients, so it must have a real root. I didn't want to sound stupid and I said sure.

I can't figure out if he's right. Is he right? Can someone help me with this?

Edit: Earlier I had asked if it must have a negative root. It's the real root that he said such polynomial must have.

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    $\begingroup$ Consider the 3rd degree polynomial given by expanding out $(x-1)(x-2)(x-3)$. This polynomial has no negative roots. $\endgroup$ – tomcuchta Jun 19 '12 at 23:35
  • $\begingroup$ What is true is that a cubic has to have at least one real root. $\endgroup$ – Dylan Moreland Jun 19 '12 at 23:38
  • $\begingroup$ Thanks for the answers. As @alex.jordan replied to the question below, I realized it's what my friend said. He didn't say "negative root", he said "real root". I'll edit my question. Sorry for confusion! $\endgroup$ – bodacydo Jun 19 '12 at 23:39
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It is true that a cubic polynomial must have a real root. Since the lead coefficient is not $0$, we have that $$ \lim_{x\to-\infty}ax^3+bx^2+cx+d=\left\{\begin{array}{}-\infty&\text{if }a>0\\+\infty&\text{if }a<0\end{array}\right. $$ and $$ \lim_{x\to+\infty}ax^3+bx^2+cx+d=\left\{\begin{array}{}+\infty&\text{if }a>0\\-\infty&\text{if }a<0\end{array}\right. $$ Since a polynomial is continuous, by the Intermediate Value Theorem, if it takes a positive value and a negative value, it must take every value in between, in particular $0$.

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If $p(x) = ax^3 + bx^2 + cx + d $ and if $a$ and $d$ happen to be of same sign, then you can conclude that there must be at least one negative root. Probably this is what your friend would have done as well.

EDIT

The question was changed from having a negative root to there is at least one real root.

This is a consequence of the fact that $p(x)$ is continuous and the intermediate value theorem for continuous functions.

The important observation is that $$\lim_{x \rightarrow \infty} p(x) = \text{sign(a)} \cdot \infty$$ and $$\lim_{x \rightarrow -\infty} p(x) = -\text{sign(a)} \cdot \infty$$

For instance, if $a > 0$, then $\displaystyle \lim_{x \rightarrow \infty} p(x) = \infty$ and $\displaystyle \lim_{x \rightarrow -\infty} p(x) = -\infty$.

Since the function is continuous, by intermediate value theorem, it must hit all values between $- \infty$ and $\infty$ for some $x$. Hence, in particular, it must hit $0$. Intutively, the function is positive for some large $x$ and negative for some large $x$ and since the function has no jumps, it must hit $0$ somewhere on the real axis before it changes sign.

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  • $\begingroup$ my friend actually said "real root" not "negative root". I mixed the terms up as I was recalling what he said. I'm sorry about that. I edited my question. $\endgroup$ – bodacydo Jun 19 '12 at 23:41
  • $\begingroup$ Just to elaborate on Marvis' observation, a monic cubic will factor, at least over $\mathbf C$, as $(x - r_1)(x - r_2)(x - r_3)$. So the constant term is $-r_1r_2r_3$. Now investigate the case in which this term is $> 0$. [I don't see a way of avoiding splitting this up into a case for three real roots and another for two complex conjugate roots.] $\endgroup$ – Dylan Moreland Jun 19 '12 at 23:48
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    $\begingroup$ @DylanMoreland Another way to look at it is that $a$ and $d$ are of same sign is equivalent to saying that $p(0)$ and $p(\infty)$ will have the same sign. Hence, there are either two roots (I include double roots as two roots) in $[0, \infty)$ or no roots in $[0, \infty)$. $\endgroup$ – user17762 Jun 19 '12 at 23:55
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Any polynomial $p(x)$ of odd degree satisfies $$ \lim_{x \rightarrow \infty} p(x) = \pm \infty $$ and $$ \lim_{x \rightarrow -\infty} p(x) = \mp \infty. $$ That is, one side of the graph goes up forever and the other side of the graph goes down forever. Since polynomials are also continuous, the graph has no choice but to cross the $x$ axis somewhere, giving a real root. (The fact that the continuous function has "no choice" is formalized by the Intermediate Value Theorem).

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    $\begingroup$ Wow, that is a very nice argument. Thanks! $\endgroup$ – bodacydo Jun 19 '12 at 23:43
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No, cubic polynomials must have a real root. But it does not have to be negative. Consider $x^3-1$. Its only real root is the positive number $1$.

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  • $\begingroup$ Oops, that's what my friend actually said! He didn't say "negative root", he said "real root". Now just how do you prove that exactly... $\endgroup$ – bodacydo Jun 19 '12 at 23:38
  • $\begingroup$ @AustinMohr Thanks, I'm editing it right now. $\endgroup$ – bodacydo Jun 19 '12 at 23:39
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    $\begingroup$ Complex roots always come in conjugate pairs when you have real coefficients. Think about why that is. $\endgroup$ – tomcuchta Jun 19 '12 at 23:43
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Other answers used the intermediate value theorem. Here's an alternative. By the complex conjugate root theorem, non-real roots occur in complex conjugate pairs. By the fundamental theorem of algebra, a cubic polynomial has three complex roots. It follows that a real cubic polynomial must have at least one real root.

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