Question

State the number of complex zeros, the possible number of real and imaginary zeros, the possible number of positive and negative zeros, and the possible rational zeros for each function.

Answer 1

. one real zero, and 4 complex zeros , 1 rational zero, 3 negative zeros, 2positive 0

Answer 2

?

Answer 3

haven't done this in a while, a little rusty

the degree (highest exponent) is 5, so 5 possible zeros

you can look at the number of sign changes in f(x) to get the # of positive real zeroes. since everything is positive, there are zero sign changes and thus zero positive real zeroes.

plug in -x into f(x), count the number of sign changes again to get the number of possible negative real zeros

# of complex zeroes is simply 5 - (# of real zeroes)

possible rational zeroes can be calculated through the rational root theorem

Answer 4

So everything is right except the possible number of rational roots is +/-1, +/-1/3+/-3+/-9

Answer 5

ah, I didn't see your answer, I can check real quick

Answer 6

is this 10th-grade math

Answer 7

algebra

Answer 8

oh its hard

Answer 9

I think you're missing a few for rational zeroes

it wants all possible rational zeroes, so *every factor* of 9 over *every factor* of 15

factors of 9: 1, 3, 9
factors of 15: 1, 3, 5, 15

so +/- 1/1, +/- 3/1, +/- 9/1
+/- 1/3, +/- 3/3, +/- 9/3
+/- 1/5, +/- 3/5, +/- 9/5
+/- 1/15, +/- 3/15, +/- 9/15

of course you can remove duplicates

Answer 10

for negative real zeroes, you have 5 sign changes

f(-x) = -15x^5 + 3x^4 - 140x^3 + 28x^2 - 45x + 9

(negative to positive, to negative, to positive, to negative, to positive again)

you could have up to 5 negative real zeroes, but remember, since complex roots come in pairs, you can subtract 2 roots at a time to get 5, 3, or 1 negative real root)

Answer 11

in theory you would test all the positive and negative real roots, but to quicken things up, I used a calculator to get x = -1/5 as the one negative real root, meaning (x+1/5) = 0 is one of the roots.

when you factor out (x+1/5) (in theory, using synthetic division) you get 15x^4 + 140x^2 + 45 which factors to 5(3x^4 + 28x^2 + 9) and again to 5(x^2+9)(3x^2 + 1)

both x^2 + 9 and 3x^2 + 1 have two complex root solutions, giving us 4 complex roots

Answer 12

State the number of complex zeros ---> 4
the possible number of real and imaginary zeros ---> complex zeroes come in pairs. real + complex have to add up to 5. we have 5, 3, or 1 negative real zeroes based on descarte's sign change theorem, so our possibilities are:

5 negative real zeroes, 0 complex zeroes
3 negative real zeroes, 2 complex zeros
1 negative real zeroes, 4 complex zeros

the possible number of positive and negative zeros ---> 0 positive. 5, 3, or 1 negative zero

the possible rational zeros ---> so +/- 1/1, +/- 3/1, +/- 9/1
+/- 1/3, +/- 3/3, +/- 9/3
+/- 1/5, +/- 3/5, +/- 9/5
+/- 1/15, +/- 3/15, +/- 9/15 (remove the duplicates)

Answer 13

ok, solving the polynomial was probably a little overkill, but I hope you get the idea

Answer 14

It definitely is, and yes thank you very much I sort of did the same way you did except stupid errors, thank you