Question

What are the possible number of positive real, negative real, and complex zeros of
f(x) = –7x^4 – 12x^3 + 9x^2 – 17x + 3?


Positive Real: 4, 2 or 0
Negative Real: 1
Complex: 0 or 1 or 3

Positive Real: 3 or 1
Negative Real: 2 or 0
Complex: 1

Positive Real: 1
Negative Real: 3 or 1
Complex: 2 or 0

Positive Real: 3 or 1
Negative Real: 1
Complex: 2 or 0

Answer 1

there are 4 zeros all together, but they don't need to be all real. the coefficients are
-7, -12, 9, -17, 3
there are 3 changes of sign, from -12 to 9, 9 to -17, 17 to 3

descartes rule of sign says that therefore there are either 3 positive zeros or 1 positive zeros

Answer 2

so you can discard answer 1 and 3. now to find the possible negative real zeros compute
\[f(-x) = –7x^4 + 12x^3 + 9x^2 + 17x + 3\]

Answer 3

now there is only one change in sign, from -7 to 12, meaning there is at most 1 negative 0

Answer 4

so the last answer is correct. note that the complex zeros come in conjugate pairs, so you cannot have 1, or 3 of them. either there are none or two

Answer 5

Thank you so much, this has really helped me.