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Suppose I have a equation of a degree of 4 and I don't know a proper method of solving this type of equation (like completing the square is a proper method to solve the quadratic equation) so how or what necessary steps should I follow so that I could guess more precisely the roots of such equation.

I asked this question because I encountered a problem in a competitive exam which asked to tell the number of distinct real roots of a equation of degree 4.

the question being:

the number of distinct real roots of $x^4-4x^3+12x^2+x-1=0 $ is $\_\_\_\_\_$

We all know that a polynomial of degree n can have maximum n roots. So the above equation can have maximum 4 roots. So I wrote 4 as the answer since there was no negative marking but I checked out by making the graph of this equation in a graph calculator which showed that this equation has 2 roots. So if would know how to find out the roots then it could reward me more marks.

So what should be done?

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    $\begingroup$ If we just want the number of real roots of a polynomial with real coefficients, there are methods that don't exactly require finding the roots. You can use Cartesian rule of signs to get an upper bound on the number of real roots. An even stronger method is to use Sturm's theorem to get the exact number of roots as well as their position (in terms of a real interval wherein it lies). $\endgroup$ – learner May 4 '16 at 13:43
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If a degree 4 polynomial has 4 real roots, then it must have at least 3 local extremes, so its derivative must have 3 real roots; by repeating this argument, its second derivative must have 2 real roots. But in fact, the second derivative of this function is $$ 12((x-1)^2+1), $$ which is clearly positive everywhere.

Edit (thank you almagest): This argument gives the upper bound on the number of roots. Finishing the argument requires inspecting the function to demonstrate that it has at least two roots, e.g. at evaluating the function at -1, 0 and 1.

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    $\begingroup$ So it has 0,1 or 2 real roots. Then $f(-1)>0,f(0)<0,f(1)>0$ so it has 2. $\endgroup$ – almagest May 4 '16 at 13:47
  • $\begingroup$ I was wondering if the following is a formal fact: "if a polynomial has N extremes, then it must have N+1 Real roots"? This has come to mind when trying to solve this question: math.stackexchange.com/questions/3319821/… $\endgroup$ – NoChance Aug 11 at 12:58
  • $\begingroup$ @NoChance no, certainly not. $x^2 +1$ is a counterexample. $\endgroup$ – Mees de Vries Aug 11 at 17:36
  • $\begingroup$ Thank you for your reply. $\endgroup$ – NoChance Aug 12 at 3:05
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This is probably not the best answer, but this is how I would have done it, since I guess in this kind of exam time mathers?

First let $f(x) = x^4 -4x^3 +12x^2 + x -1$ then I would noticed that $f(0) = -1$. After that I would look at $f'$

We have that $f'(x) = 4x^3 - 12x^2 + 24x + 1$, then one can notice that $$ 4x^3 - 12x^2 + 24x > 0 \ \forall x > 0 \Rightarrow f'(x) > 0 \ \forall x > 0$$ So $f$ is increasing for $x>0$.

At the same time on can notice that $$4x^3 - 12x^2 + 24x < 0 \ \forall x < 0$$

But here it is a bit tougher, the $+1$ will imply that $f'(x) < 0 \ \forall x< 0 - \lambda$. Where $\lambda$ is the absolutevalue of the solution to $$ 4x^3 - 12x^2 + 24x + 1 > 0$$. So the $+1$ is actually only a movement right to left in the graph.

So the conclusion is that $f'(x) > 0 \ \forall x > -\lambda, \ f'(x) < 0 \ \forall x < -\lambda$. Hence by logic and theorems, we have only two reell distincts solutions to $f(x) = 0$ The rest are complex, but are they disctinct?

I am aware that this is not properly, but as I said before, I guess you wanted to answer the question as fast as possible.

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In fact, you could go crazy and use Ferrari's solution and try to get something out of it (https://en.wikipedia.org/wiki/Quartic_function#Ferrari.27s_solution).

But this question just want you to be a little bit clever. Let $$P(x):=x^4 - 4x^3 +12 x^2 + x - 1.$$ We look at

\begin{cases} % \lim\limits_{z\to-\infty} P(z) &= +\infty.\\ P(-1) &= 15 &>0, \\ P(0) &= -1 & <0, \\ P(1) &= 9 &>0.\\ % \lim\limits_{z\to+\infty} P(z) &= +\infty. \end{cases}

Moreover, given \begin{align} P^\prime(x):=\frac{\mathrm{d}}{\mathrm{d}x} P(x)=4 x^3-12 x^2+24 x+1 \quad \text{and} \quad P^{\prime\prime}(x):=\frac{\mathrm{d^2}}{\mathrm{d}x^2} P(x)=12 x^2 - 24 x^2 +24, \end{align} we can clearly see that $P^{\prime\prime}(x) > 0 \; \forall x \in \mathbb{R}$, and \begin{cases} P^{\prime}(-1) &= -39 &<0, \\ P^{\prime}(17) &= 17 &>0. \end{cases} thus $P(x)$ is convex, with two real roots are in one $(-1,0)$ and the other one in $(0,1)$.

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Okay, this maybe a moot point at this time, but here is the two step technique I would use to "find the number of distinct roots". This is NOT an attempt to find what they are.

  1. Descartes' Sign Change Rule

    y(x): {1, -4, 12, 1, -1}  --> 3 sign changes
    y(-x): {1, 4, 12, -1, -1} --> 1 sign change
    

Possible Scenarios

    Pos.     Neg.     Cmp. Cnj. Pair     Total
     3        1             0              4
     2        1             1              4
  1. Find the sign of the discriminant of the quartic polynomial.

This involves formula with 16 terms. To abbreviate: discriminant is a negative value; therefore, option 2 is the correct solution.

There are two distinct roots and one complex conjugate pair.

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