How many roots of $x^4-4x^2-8x+12$ lie in the range $[-2,2]$

Question

I tried to use the Sturm Theorem and computed the Sturm sequence as the following: $$ P_0 = x^4-4x^2-8x+12 $$ $$ P_1 = 4x^3-8x-8 $$ $$ P_2 = 3x^3+2x^2-18 $$ $$ P_3 = 8x^2/3+8x-16 $$ $$ P_4 = 60-39x $$ $$ P_5 = something\ positive $$ where the sequence gives $V(-2)=2$ as $(+,-,-,-,+,+)$ and $V(2)=3$ as $(-,+,+,+,-,+)$, which gives $V(-2)-V(2)=-1$. It took me so much time evaluating the sequence, and it didn't even give me the correct answer somehow. I did redo the calculation, and it seemed to be correct. If someone can find the error in my calculation, please do. Otherwise, I am looking for another easier and simpler method to solve this type of question.

Answer 1

I don't know the method you mentioned, but I solve these problems by (1) drawing a rough graph of the polynomial, especially its turning points, and (2) using intermediate value theorem.

First, we have $f(x) = x^4 - 4x^2 - 8x + 12$, so $f'(x) = 4x^3 - 8x - 8 = 4(x^3 - 2x - 2)$. Note that $f'(1) = -12 < 0$ and $f'(2) = 8 > 0$, so there is a turning point in $[1, 2]$. Next, I try to determine which of the two cases $f'(x)$ is:

To do so, I find the local maxima of $f'(x)$ by computing $f''(x) = 4(3x^2 - 2)$ and evaluating $f'\left(-\sqrt{\frac{2}{3}}\right) \approx f'(-0.8) \approx 4 \cdot (-0.512 + 1.6 - 2) < 0$, so it is the case on the left. This means that $f(x)$ only has a single turning point somewhere in $[1, 2]$, and we can draw this graph:

What this means is that we can split the real line into $(-\infty, 1.\ldots)$ and $(1.\ldots, \infty)$, and $f$ is decreasing in the first and increasing in the second. We can check whether there are any real roots in the two intervals using the intermediate value theorem!

Note that:

• $f(-2) = 16 - 16 + 16 + 12 = 28 > 0$
• $f(1) = 1 - 4 - 8 + 12 = 1 > 0$
• $f(2) = 16 - 16 - 16 + 12 = -4 < 0$

The latter two equations show that $f(1.\ldots) < 0$, and also that the root on $(1.\ldots, \infty)$ is outside $[-2, 2]$. On the other hand, $f(-2)f(1.\ldots) < 0$ implies that there is a single real root in $[-2, 1.\ldots)$.

In the end, there are $\fbox{1}$ roots between $[-2, 2]$.

Answer 2

With Sturm's theorem: Note that we could, at any stage, replace $P_i$ with $\tilde{P_i}=\lambda P_i$ where $\lambda>0$. But for the sake of conforming, \begin{align*} P_0(x)&=x^4-4x^2-8x+12\\ P_1(x)&=4x^3-8x-8\\ P_2(x)&=-\operatorname{rem}(P_0,P_1)=2x^2+6x-12\\ P_3(x)&=-\operatorname{rem}(P_1,P_2)=-52x+80\\ P_4(x)&=-\operatorname{rem}(P_2,P_3)=-P_2(\frac{80}{52})<0 \end{align*} So $V(-2)=3$ and $V(+2)=2$ (the sequences are $(+,-,-,+,-)$ and $(-,+,+,-,-)$), giving one root in $[-2,2]$.

Alternative method: if you know the discriminant of a polynomial, then a simple calculation shows $$ \operatorname{disc}P=\operatorname{Resultant}(P,P')=-339968 $$ where the first equality is from $P$ being a monic quartic and the second is explicit calculation of the determinant. So $P$ has two real roots and two nonreal roots. Hence the change of sign $P(2)<0<P(-2)$ gives the one real root in $[-2,2]$.

Answer 3

Another way:

The equation $x^4-4x^2-8x+12=0$ can be written as $$(x^2-2)^2=8(x-1).$$ From the table below and the intermediate-value theorem, we deduce that there are two roots of the equation, one in the interval $(1,2)$, the other in the interval $(2,3)$. $$ \begin{array}{c|c|c} x & (x^2-2)^2&8(x-1) \\ \hline 1 & 1&0\\ \hline 2 & 4&8 \\ \hline 3&49&16\\ \hline \end{array} $$ To show that there are exactly two real roots we need Descartes'rule of signs and graphical support: Signs of the non-zero term coefficients in order are $+1,-4,-8,+12$. There are two sign changes: One from $+1$ to $-4$, the other from $-8$ to $+12$. Therefore there are at most two positive roots. From graph, we see that there are no negative roots. Therefore our equation has two positive real roots and two complex conjugate roots.