Determine a possible quartic polynomial equation such that $f(x) > 0$ for $-4 < x < -2$ and $3 < x < 7$

Question

In typical high school fashion, nowhere in the curriculum was there a question about creating a polynomial equation. Yet here it is in the exam practice questions. Little help please?

Answer 1

A polynomial will change sign at the roots.

If the roots are $-2,-4, 3, 7$ then $P(x) = a(x+2)(x+4)(x-3)(x-7)$ and for $x\le 3$ we have $x+2,x+4, x-3, x-7 < 0$ and $P(x) \le 0$ so $a < 0$ and for $-2 < x < -4$ we have $(x+2) > 0, x+4 < 0, x-3< 0, x-7 < 0$ so $a < 0$ means $P(x) > 0$ and so on.

SO $P(x) = -(x+4)(x+2)(x-3)(x-7)$ is one possible solution.

But the question never said $f(x) \le 0$ outside the intervals. So long as the roots (if any) are outside the intervals and there are the right even or odd number of roots between $-2$ and $3$ it will work.

$P(x) = (x+1)x(x-1)(x-2)$ will be acceptable as $x+1,x,x-1, x-2 < 1$ for $-4 < x < -2$ so $P(x) > 0$ there and $x+1,x, x-1, x-2 > 0$ for $3< x< 7$ so $P(x) > 0$ there.

We can even do $P(x) = x^4 + 1$ as that is positive anywhere. (and has no roots).

Answer 2

The question only makes sense if $f(x)>0$ only in the given range. Then by continuity, $f(x)$ must be $0$ at $x=-4,-2,3,7$, and we take

$$f(x)=\pm(x+4)(x+2)(x-3)(x-7),$$ which is indeed a quartic polynomial.

The sign alternates between the roots so taking the $-$ sign fulfills the request.