Question

State how many imaginary and real zeros the function has.

f(x) = x3 + 5x2 + x + 5

Answer 1

are you familiar with descarte's rule of signs?

Answer 2

yea

Answer 3

use it

Answer 4

positive: 0
negative: 3

Answer 5

that tells me how many roots but it doesnt tell me how many are imaginary or real

Answer 6

hmm i dont know much about this anymore so ill let @sami-21 take over to avoid mistakes

Answer 7

ok let me try.
there is no positive root because no change of sign
now try this.
\[\Large f(-x)=-x^3+5x^2-x+5\]
since there three change of signs . so three negative roots.

Answer 8

yea i got that but the question is asking how many imaginary and real zeros are there and i dont know how to determine if the zeros are imaginary or real after that point

Answer 9

i would just try to find the zeros normally right?

Answer 10

descarte rules gives you information about the positive negative and imaginary roots.
if a polynomial of degree n has P positive roots q negative roots then it has at least n-(p+q) imaginary roots.
since we do not have positive roots.does it mean imaginary roots are zero ????
Remember this does not give us surety that there is no imaginary root. it says the word "at Least" it will have n-(p+q) imaginary roots.
in this case it is of 3d degree polynomial so it can have at most two imaginary roots (because imaginary roots appears in pairs).

Answer 11

let me call hero .he is good at algebra stuff :)
@Hero

Answer 12

Did anyone actually try just factoring it?

When I factor I get

f(x) = x^2(x+5) + 1(x+5)

Now I set f(x) = 0, to find some roots

0 = (x+5)(x^2 + 1)