Question

Use the Rational Zeros Theorem to write a list of all possible rational zeros of the function.

f(x) = -2x4 + 4x3 + 3x2 + 18

Answer 1

Please Help ASAP!!

Answer 2

use the thrm ... to list the possible ... it tells you exactly what to do :/

Answer 3

But I don't know how to do it

Answer 4

tell me what the thrm states ...

Answer 5

If P(x) is a polynomial with integer coefficients and if is a zero of P(x) ( P() = 0 ), then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x) .

Answer 6

good ... seems a little incomplete but lets say it like this:

we have a poly of the form:\[P(x)=ax^n+bx^{n-1}+...+jx+k\]
IF there exists a rational root to this poly, then it will have to be of the form:\[\pm\frac{factors~of~k}{factors~of~a}\]

Answer 7

what are the factors of the constant: 18
what are the factors of leading coeff: 2

Answer 8

Factors of 2 are: 2*1
Factors of 18 are:18*1, 2*9, 3*6,

Answer 9

very good now lets see how we can pair them up:\[\pm\frac{1,2,3,6,9,18}{1,2}\]
\[\pm\frac{1}{1},\pm\frac{2}{1},\pm\frac{3}{1},...,\pm\frac{18}{1}\]
\[\pm\frac{1}{2},\pm\frac{2}{2},\pm\frac{3}{2},...,\pm\frac{18}{2}\]
reduce and remove duplicates to simplify the list.

Answer 10

show me your final list, or ask me questions if its unclear :)

Answer 11

you can just remove 2/1 and 2/2 since they are the same

Answer 12

2/1 does not equal 2/2 .... i believe what your thought was: 1/1 = 2/2: 1=1. having two 1s in the list is redundant.

Answer 13

2/2, 6/2, and 18/2 create duplicate elements, but the other /2 elements are part of the list if i see it right

Answer 14

Ok

Answer 15

to compare, that leaves us:

+- (1,2,3,6,9,18, 1/2, 3/2, 9/2) as all the potential rational roots

Answer 16

ok

Answer 17

does it make more sense now?

Answer 18

Yes

Answer 19

good luck :)