Question

1

Answer 1

and the rational zeros theroem

Answer 2

if you just want to know the possible combinations of positive/negative/complex solutions, use descartes rule of signs by itself

Answer 3

so that means there are either 3 positive real zeros or 1 positive real zero

Answer 4

you must analyze f(-x) for possible negative real zeros

Answer 5

no, plug in -x for x

Answer 6

f(x) = 7x^5 + 15x^4 – x^3 + 4x^2 – 6x – 11
f(-x)=7(-x)^5+15(-x)^4-(-x)^3+4(-x)^2-6(-x)-11
=-7x^5+15x^4+x^3+4x^2+6x-11

Answer 7

2 sign changes in f(-x) so either 2 or 0 real negative zeros

Answer 8

so make a table with possible combinations of zeros

Answer 9

positive 3 3 1 1
negative 2 0 2 0
complex 0 2 2 4

Answer 10

each column must add to 5 because it is a 5th degree polynomial with 5 zeros

Answer 11

if there are 3 positive real zeros, there can be either 2 or 0 negative real zeros

Answer 12

if there are 3 positive real zeros and 0 negative real zeros, the rest must be complex conjugates

Answer 13

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