Question

please help!!! my answer is C

What are the possible number of positive, negative, and complex zeros of f(x) = -2x3 + 5x2 - 6x + 4 ? (3 points)
Answer

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

Positive: 1; Negative: 2 or 0; Complex 2 or 0

Positive: 2 or 0; Negative: 1; Complex: 2 or 0

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

Answer 1

Sorry, that's not correct...

Answer 2

Do you know about Descartes' Rule of Signs?

Answer 3

Count up the number sign changes in the coefficients of the polynomial. Here we have 3. That means (by the RoS) that we have either 1 or 3 positive roots.

Answer 4

Next, figure out what the polynomial would be if we rewrote it in terms of -x. Essentially, if you have an odd power of x, change the sign, otherwise leave it alone. That gives us
+2x^3 + 5x^2 + 6x + 4. There are no sign changes there, so that means we have 0 negative roots. We know by the rational root theorem that we have a root for each power of x (3 in this case, because x^3 is the highest order term) so we must have either 3 positive roots, or 1 positive root and 2 complex roots (complex roots always come in conjugate pairs when we have a polynomial with real coefficients).

Answer 5

So, count the sign changes in P(x). That will be the number of positive roots, minus a multiple of 2.

Next, count the sign changes in P(-x). That will be the number of negative roots, again minus a multiple of 2.

In our case, here our count of sign changes in P(x) was 3, and the multiples of 2 less than 3 are 0 and 2, so we can have either 3-0 = 3 or 3 - 2 = 1 positive roots. Our count of sign changes in P(-x) was 0, so 0 negative roots.