Can all the roots of $ax^5+bx^2+c=0$, with real coefficients and $a,c\neq0$, be real numbers?

Question

Let $ax^5+bx^2+c=0$ and $a,c\neq 0$ and $a,b,c$ are real numbers. Can all the roots of this quintic equation be real numbers?

I divided each side by $a$ and I got

$$x^5+\frac bax^2+\frac ca=0$$

Using trial and error, I found at most $3$ roots are real numbers.For example,

$$x^5+7x^2-3=0$$

I couldn't find the case, so that $5$ roots are real numbers.

Answer 1

This answer assumes $a,b,c$ are real numbers. [If they can be arbitrary complex nothing can be said.] Using DeCarte's rule of signs, it can have at most two positive zeroes, and also at most two negative zeroes. Since you assume $a,c \neq 0$ it does not admit zero itself as a zero. Simply using the DeCarte's rule this gives at most 4, however it cannot have exactly 4 since the complex roots come in conjugate pairs.

Answer 2

The derivative has two roots, so the function has two extrema, so it has at most three roots.