# Counting the Number of Real Roots of $y^{3}-3y+1$

Here's my question:

How many real roots does the cubic equation $y^3-3y +1$ have?

I graphed the function and it crossed the x-axis $3$ times. But my professor doesn't want a graphical explanation. So in that case, I was looking at the Fundamental Theorem of Algebra and states that a polynomial of degree n can have at most n distinct real roots. so therefore, there must be 3 real roots?

EDIT

It seems that there are numerous ways to approach this problem after all. And we can expand this to other types of polynomials as well, not just cubics.

• The cubic equation $y = x^3 + 1$ only has one real root. – Travis Willse Mar 13 '17 at 17:59
• At any rate, this can be determined using the discriminant of the polynomial. mathworld.wolfram.com/PolynomialDiscriminant.html – Travis Willse Mar 13 '17 at 17:59
• The fundamental theorem only gives an upper bound on the number of roots. But you can make your "graphical" argument mathematically correct by picking suitable points above and below the $x$ axis and then explicitly evaluating the polynomial at those points. – David K Mar 13 '17 at 18:01
• In short, yes, see en.wikipedia.org/wiki/Discriminant#Degree_3 . – Travis Willse Mar 13 '17 at 18:36
• It's not an equation unless you set it equal to 0 (or some other expression): $y^3−3y+1 = 0$ – smci Mar 14 '17 at 9:46

Nobody's explicitly written out the discriminant-based solution.

The discriminant of the polynomial $x^3 + px + q$ is $-4p^3 - 27q^2$ - see Wikipedia on discriminant of third-degree polynomials. So $y^3 - 3y + 1$ has discriminant $-4 \times (-3)^3 - 27 \times 1^2 = +81$. Since this is positive, the polynomial has three distinct real roots.

The given polynomial evaluated at $y\in\{-2,0,1,2\}$ exhibits three sign changes, hence it has at least $3$ real roots, and obviously cannot have more than three roots.

• Probably not the point of the question, but the roots are $2 \cos \left( \frac{2 \pi}{9} \right) \approx 1.532, \; \; 2 \cos \left( \frac{4 \pi}{9} \right) \approx 0.347, \; \; 2 \cos \left( \frac{8 \pi}{9} \right) \approx -1.879.$ – Will Jagy Mar 13 '17 at 18:40
• @WillJagy: that's an interesting addendum. – Jack D'Aurizio Mar 13 '17 at 18:45
• @unseen_rider: See Viète's construction of the roots of the cubic. (A derivation is further up in that same article.) Note that the middle root can be stated a bit more nicely as as $2 \cos \left( \frac{14 \pi}{9} \right)$, which puts the roots evenly spaced around a unit circle. – Michael Seifert Mar 13 '17 at 19:06
• @MichaelSeifert I don't think it follows from Viete. This is from a method of Gauss. I found a discussion in Galois Theory by David A. Cox. Then many explicit examples in Reuschle (1875). In any case, I have added a proof to my answer. – Will Jagy Mar 13 '17 at 19:15
• @WillJagy: Ah, you came at it from a completely different direction than I did. I've added an answer showing how you can also use Viète's construction to get the same answer as you did. – Michael Seifert Mar 13 '17 at 19:27

The function has extrema where

$$3y^2-3=0$$ i.e. at $$y=\pm1$$. The values at these extrema are $$3$$ and $$-1$$.

So the variations of this continuous function are $$-\infty,3,-1,\infty$$, proving that there are three changes of sign.

For a cubic polynomial, the search for the extrema may be more efficient than trial-and-error because it is immediately conclusive.

In the case of a depleted polynomial $$x^3+px+q$$

the extrema are located at $$x=\pm\sqrt{-\frac p3}$$ if $$p<0$$.

In this case there are three real roots if

$$\sqrt{-\frac p3}^3-p\sqrt{-\frac p3}+q>0\text{ and }-\sqrt{-\frac p3}^3+p\sqrt{-\frac p3}+q<0,$$ or

$$\frac{2p}3\sqrt{-\frac p3}<-|q|.$$

• Probably not the point of the question, but the roots are $2 \cos \left( \frac{2 \pi}{9} \right) \approx 1.532, \; \; 2 \cos \left( \frac{4 \pi}{9} \right) \approx 0.347, \; \; 2 \cos \left( \frac{8 \pi}{9} \right) \approx -1.879.$ – Will Jagy Mar 13 '17 at 18:40
• @ Will Jagy Well, it's good to know the solutions either way. – John Bradshaw Mar 13 '17 at 18:43

Let $f(y) = y^3 - 3y + 1$. Then you can observe that $f(0) = 1, f(1) = -1$; thus $f$ has at least one root between $0$ and $1$, by the intermediate value theorem. You can find the other two roots similarly.

You could use Cardano's Formula to verify it explicitly.

Perhaps a kind of cool way that's completely overkill:

Look at the polynomial $P(y) = \frac{1}{4}y^4 - \frac{3}{2}y^2 + y$, whose derivative is $y^3 - 3y^2 +1$

You can factor $P(y) = \frac{1}{4}y(y-2)(y^2+2y-2)$. use the quadratic formula to determine that the roots of the third factor are real.

By the Gauss Lucas-Theorem (the roots of the derivative of $P$ are contained in the convex hull of the roots of $P$), the roots of $P'(y)$ must also be real, and $P'$ is the polynomial you're interested in.

• Really nice (and a very unlikely approach for a beginning student!). – John Hughes Mar 13 '17 at 18:57

Three. The roots are $$2 \cos \left( \frac{2 \pi}{9} \right) \approx 1.532,$$ $$2 \cos \left( \frac{4 \pi}{9} \right) \approx 0.347,$$ $$2 \cos \left( \frac{8 \pi}{9} \right) \approx -1.879.$$

See page 174 in Reuschle. I first came across the method of Gauss in chapter 9 of Galois Theory by David A. Cox. Chapter 9 is called Cyclotomic Extensions. Section 9.2 is called Gauss and Roots of Unity. I then found a brief mention of the 1875 Reuschle book on page 195 of Theory of Numbers by Mathews (1892). He wrote

The reader who wishes for more numerical illustrations should consult Reuschle's Tafeln Complexer Primzahlen (Berlin, 1875).

Or, let $\omega$ be a primitive 9th root of unity, so $$\omega^9 = 1,$$ $$\omega^3 \neq 1.$$ For $$t^9 - 1 = (t-1)(t^2 + t + 1)(t^6 + t^3 + 1),$$ we find $$\omega \neq 1, \; \; \omega^2 + \omega + 1 \neq 0, \; \; \omega^6 + \omega^3 + 1 = 0.$$

Then take $$x = \omega + \frac{1}{\omega}$$ and calculate $$x^3 - 3 x + 1 = \omega^3 + 1 + \frac{1}{\omega^3} = \frac{\omega^6 + \omega^3 + 1}{\omega^3}$$

• @ Will Jagy How did you compute the roots? – John Bradshaw Mar 13 '17 at 18:11
• @i I looked it up in Reuschle (1875) page 174. Easy to prove. – Will Jagy Mar 13 '17 at 18:14

Just for fun: we note that since we have a "depressed cubic" of the form $t^3 + pt + q = 0$ (i.e., no quadratic term), the roots have a nice geometric interpretation: $$t_k = 2 \sqrt{ -\frac{p}{3}} \cos \left( \frac{1}{3} \arccos \left( \frac{3q}{2p}\sqrt{ - \frac{3}{p} } \right) + \frac{2 \pi k}{3} \right)$$ for $k = {}$0, 1, or 2. These can be seen to be the $x$-coordinates of three points equally spaced around the unit circle. In the given case, we have $p = - 3$ and $q = 1$, so this reduces to $$t_k = 2 \cos \left( \frac{1}{3} \arccos\left( -\frac{1}{2} \right) + \frac{2 \pi k}{3} \right) = \left\{ 2 \cos \left( \frac{2\pi}{9} \right), 2 \cos \left( \frac{8\pi}{9} \right), 2 \cos \left( \frac{14\pi}{9} \right) \right\}.$$ Note that $\cos (14\pi/9) = \cos (4\pi/9)$, so this agrees with Will Jagy's answer as well.

• I had not known one could go this root. Or route. – Will Jagy Mar 13 '17 at 19:31
• Michael, as you know Viete's method, can he get $x^3 + x^2 - 4x + 1?$ Reuschle page 15. Each root needs two cosine terms, denominators 13. – Will Jagy Mar 13 '17 at 20:02
• @WillJagy: I don't have Reuschle's book, but my guess is that it yields the roots in a different form. In general, Viète's method yields the three roots in the form $x_i = \bar{x} + A \cos (\theta_i)$, where $\bar{x}$ is the average of the three roots (which is equal to 1/3 of the coefficient of the quadratic term), $A$ is a real number, and the $\theta_i$ are three angles equally spaced around the unit circle. In this case, $\bar{x} = 1/3$; and since $\cos \theta \neq 1/3$ for any rational angle, it would take some work to show that these two forms are equivalent. – Michael Seifert Mar 13 '17 at 21:01
• In case of interest: the roots of $x^3 + x^2 - 4 x + 1$ are $$2 \cos \left( \frac{2 \pi}{13} \right) + 2 \cos \left( \frac{10 \pi}{13} \right) \approx 0.27389, \; \;$$ $$2 \cos \left( \frac{4 \pi}{13} \right) + 2 \cos \left( \frac{6 \pi}{13} \right) \approx 1.37720, \; \;$$ $$2 \cos \left( \frac{8 \pi}{13} \right) + 2 \cos \left( \frac{12 \pi}{13} \right) \approx -2.65109 \; \;$$ – Will Jagy Mar 13 '17 at 22:20
• frm page 15 of online books.google.com/… For example the first root is given as $$\alpha + \alpha^5 + \alpha^{-5} + \alpha^{-1}$$ where $\alpha = \exp (2 \pi i / 13)$ – Will Jagy Mar 13 '17 at 22:23

let $f(y)=y^3-3y+1$ then $f'(y)=3y^2-3$ and we have $$f'(y)=0$$ if $$y\pm1$$ since $$f''(y)=6y$$ we have $$f''(1)=6>0$$ and $$f''(-1)=-6<0$$ thus we have for $y=1$ a local Minimum and for $y=-1$ a local Maximum. Can you finish this?

One more interesting way to do it...

Suppose that a solution takes the form $x+iy$. Then some simple expansion and factoring gives $$(x^3-3x-3xy^2+1)+(3x^2y-y^3-3y)i=0$$ Now, if $y\neq0$ we can divide the imaginary component by $y$ to give $$3x^2-y^2-3=0$$ Substituting this into the real part gives $$8x^3=6x+1$$ Now, for $x\geq 1$, we have that $8x^3\geq8x>6x+1$, and so $x<1$. And for $x\leq-1$, we have that $8x^3\leq8x<6x+1$, so $x>-1$.

But if $-1<x<1$, then $y^2=3(x^2-1)$ is negative, which contradicts the claim that $y$ is real. Therefore, all solutions to our cubic are real (as assuming $y$ to be a nonzero real number produces a contradiction).

All that remains is to show that they are distinct - suppose that our polynomial is $(x-a)(x-b)^2$. Then expanding shows that $a=-2b$ (due to no $x^2$ term), and substituting this in gives $$(x+2b)(x^2-2bx+b^2) = x^3-3b^2x+2b^3$$ However, for this to match our polynomial, we must have $b^2=1$ and $b^3=\frac12$, a contradiction. Therefore, all roots are distinct.

Most of the answers are using the fact that it's a cubic. There is a more general way of approaching the problem which works for all polynomials.

We generate the Sturm sequence of $P(y)$: $$\begin{eqnarray} p_0(y) & = & P(y) = y^3 - 3y + 1 \\ p_1(y) & = & \textrm{content}( P'(y) ) = y^2 - 1 \\ p_2(y) & = & -\textrm{rem}(p_0, p_1) = (y^2 - 1)y - (y^3 - 3y + 1) = 2y - 1 \\ p_3(y) & = & -\textrm{rem}(p_1, p_2) = (2y - 1)\left(\frac{y}2 - \frac{1}{4}\right) - (y^2 - 1) = \frac{5}{4} \end{eqnarray}$$

We've ended up with a constant GCD, so there are no double roots.

Now we can count the real roots by looking at the number of sign changes when we evaluate the polynomials in the sequence at $\pm\infty$.

• $y = -\infty$: the signs are $-, +, -, +$ with three changes.
• $y = \infty$: the signs are $+, +, +, +$ with zero changes.

The difference in the number of sign changes is three, so there are three real roots.

Set $x=2\cos\varphi$; then the equation becomes $$4\cos^3\varphi-3\cos\varphi=-\frac{1}{2}$$ that is, $$\cos3\varphi=\cos\frac{2\pi}{3}$$ Thus we get $$3\varphi=\frac{2\pi}{3} \qquad\text{or}\qquad 3\varphi=\frac{2\pi}{3}+2\pi \qquad\text{or}\qquad 3\varphi=\frac{2\pi}{3}+4\pi$$ and therefore $$x=2\cos\frac{2\pi}{9} \qquad\text{or}\qquad x=2\cos\frac{8\pi}{9} \qquad\text{or}\qquad x=2\cos\frac{14\pi}{9}$$ Note that choosing $3\varphi=-\frac{2\pi}{3}$ wouldn't give different solutions.

• Why did you take $x= 2\cos \varphi$ ? why not something else ? – A---B Mar 15 '17 at 0:39
• @A---B The aim is making the ratio between the coefficient of the cubic and linear term to become $4/3$ – egreg Mar 15 '17 at 6:31

No. $x^3$ has only one real root, and $x^3 + x$ has only one real root and it's not a multiple root. TO show that your polynomial has more real roots, one approach is to find a value $x$ where it's positive, and a value $z > x$ where it's negative. For then between $-\infty$ and $x$ there must be a root (by the intermediate value theorem), and between $z$ and $\infty$ there must be another and ... you can probably work out the rest.

• Probably not the point of the question, but the roots are $2 \cos \left( \frac{2 \pi}{9} \right) \approx 1.532, \; \; 2 \cos \left( \frac{4 \pi}{9} \right) \approx 0.347, \; \; 2 \cos \left( \frac{8 \pi}{9} \right) \approx -1.879.$ – Will Jagy Mar 13 '17 at 18:39

As a hint, what you want to use is the related remainder theorem, which will give you that the cardinality of the set of the roots of an $n$-polynomial is $n$ for every integer $n \geq 1$. Note that the so-called fundamental theorem of algebra does not guarantee there are exactly how many roots; it guarantees there will be at most, how many, disctinct roots.

But the above is a theoretical approach that includes complex roots. To find the real roots, an observation and a decomposition will do.

We seem to be bringing out many different techniques in these answers, so here's one more: Descartes' Sign Rule.

Let $f(y) = y^3 - 3y + 1.$ There are three sign changes reading the coefficients from highest-order to lowest, so there are either two positive real roots or none. But $f(-y) = -y^3 + 3y + 1$ has just one sign change, so there is one negative real root.

Now all you have to do is exhibit the existence of at least one positive root by some other means, and then you know there must be two positive roots, which added to the single negative root gives three roots altogether.

This is not an especially helpful approach for this particular polynomial, since if you have proved the existence of at least one positive root, you can probably prove the existence of the other two roots almost immediately without using Descartes' rule.

This is an approach on how to solve the DEPRESSED cubic equation. To show the roots are real just find the discriminant of the cubic, (it can be shown easily)

Subsitute $$y= ucosa$$

Substite in the original equation to get

$$(ucosa)^3 -3ucosa+1 =0$$

Recall the identity $$cos(3a)=4cos^3a-3cosa$$

Manipulate the equation to get

$$u^3(\frac{cos3a +3cosa}{4}) -3ucosa+1 =0$$

Distribute to obtain: $$u^3(\frac{cos3a}{4})+ u^3(\frac{3cosa}{4}) -3ucosa+1 =0$$

Take the common factor out of the middle two terms to obtain:

$$u^3(\frac{cos3a}{4})+ 3ucosa(\frac{u^2}{4}-1)+1 =0$$

"u" is a parameter chosen by us. We are in search of such a parameter "$u$"; so that the middle term cancels out so that we can solve the trigonometric equation. In order for this to occur: $$u=±2$$

Substituting $u=2$ we get

$$2cos3a+1 =0$$

Solving for $cos3a$ we get

$$cos3a= \frac{-1}{2}$$

Therefore

$$3a= arccos(\frac{-1}{2})$$

$$a = \frac{arccos(\frac{-1}{2})}{3}$$

$$y= 2cos(\frac{arccos(\frac{-1}{2})}{3})$$

Similarly we can subsitute $u=-2$ and find another root.