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.