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

Question

The answer is apparently 1, but I don't know how they got it. Its from a multiple choice question where the options were: (A) 0 (B) 1 (C) 2 (D) 3 (E) 4

Answer 1

The correct answer is (B) that is one toot is between $-2$ and $2$

The potential solutions are the intersection of the two functions $$y=x^4+12$$ and $$y=4x(x+2)$$

The vertex of the quartic is $(0,12)$ and it is open upward with a sharp slope. The vertex of the quadratic is $(-1,-4)$ and it is also open upward with a smaller slope.

Note that the values of the quartic at $1,2,3$ are $13,28,93$ respectively.

The values of quadratic at $1,2,3$ are $12,32,60$ respectively.

Therefore there are two intersection points one of which is between $x=1$ and $x=2$ and the other one is between $x=2$ and $x=3$ where functions switch domination.

Answer 2

One method to count the exact number of real roots of a univariate polynomial in an interval $[a,b]$ where $a$ and $b$ can be $-\infty$ and $\infty$ respectively too, is to use Sturm's sequence. Look at this wikipedia page for a quick look. The idea is that you make a finite sequence of polynomials, $P_0$ is your original polynomial. Then $P_1$ is derivative of $P_0$ and then for any $i>1$, $P_i=-rem(P_{i-2},P_{i-1})$ which by $rem(P_{i-2},P_{i-1})$ I mean the remainder of the division of $P_{i-2}$ by $P_{i-1}$. Note that degree of the polynomials in your sequence are decreasin exactly by one in each step, so evantually you get a constant polynomial which is the step that you should stop (that will be $P_n$ where $n$ is degree of your polynomial). Then make two new sequences, the first one evaluation of your polynomial sequence at $x=a$ and the second one evaluations at $x=b$. Then consider the number of change of signs (only care about positive and negative, so ignore zeros) at the first and the second new sequences, denote them by $N(a)$ and $N(b)$, then the number of real roots in the interval $[a,b]$ is equal to $N(a)-N(b)$. You can compute by hand for your current example since it is not big, but I did it by Maple. Here is the commands I used (the commands are not in any specific package/library so you can use them without loading any package). Of course I should tell you that there are some assumptions for the Sturm's sequence such as square-free etc, but you can make your polynomial square-free by computing g.c.d. of it with its derivative etc. You can check some book or references of wikipedia to read more.

f:=x^4-4*x^2-8*x+12; # Here I defined f to be your polynomial.
sturmseq(f,x); # Here I asked Maple to show the Sturm sequence (the polynomial sequence I talked about) for f as a univariate polynomial on the variable x.
sturm(f,x,-2,2); # Here I asked Maple to check the difference of sign changes of the two new sequence of evaluations of the previous sequence at the endpoints of your interval. Of course you can ask this command without asking the previous one before it.

The result is the following;

Therefore there is exactly one root in the interval $[-2,2]$. If you want to make the two other sequences and see them yourself to check the number of sign changes, you can do as following.

L:=sturm(f,x); # Just putting a name for the polynomial sequence.
La:=eval(L,x=-2); # This makes a new list called La with entries being evaluation of elements of L at x=-2 with the same order in L.
Lb:=eval(L,x=2); # The same as La, but at x=2.

The result is as follows.

As you can see, $N(a)=3$ and $N(b)=2$, so $N(a)-N(b)=3-2=1$.