Question

y = x^{4} - x^{3} - 28x^{2} - 20x + 48 How many possible negative real zeros, positive real zeros, and non-real zeros does this equation have?

Answer 1

$f(x)= x^{4} - x^{3} - 28x^{2} - 20x + 48$

There are two changes of sign, so there are 2 or 0 possible positive roots.

$f(-x)= x^{4} +x^{3} - 28x^{2}+ 20x + 48$

There are two changes of sign, so there are 2 or 0 possible negative roots.

There are 4,2 or 0 possible non-real roots.