The number of distinct real roots of a polynomial of degree 4

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $(0,1)$.