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.
Best Answer
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.