Question

What are the possible rational zeros of f(x) = 2x3 – 4x2 – 7x + 14 ?

Answer 1

same as before, apply the Descartes signs rule to f(x) to get the POSITIVE ones
and apply it to the f( -x ) to get the NEGATIVE ones

this polynomial, like the one before is a 3rd degree polynomial, so it will have 3 roots
where the complex ones will fill in any gaps from the real roots

Answer 2

i got ± 1, ± 2, ± 7, ± 14

Answer 3

7? 14?

the polynomial is a 3rd degree polynomial, it will only have 3 roots

Answer 4

its zeros not roots

Answer 5

right? answers all have multiple numbers

Answer 6

using the rational root theorem

find the factors of the constant term call them p

so you need to find the factors of 14

+- 1, +-2, ...etc

find the factors of the coefficient of the leading term and call them q..

the leading term is 2x^3 the coefficient of 2 are +-1, +-2

the rational zeros will be p/q

Answer 7

ohh man, yes, well, so much for my eyes :(

Answer 8

so was i right or?

Answer 9

like campbell_st said, you'd need to use the rational root test, lemme check

Answer 10

yes you are correct..@spencej22

Answer 11

ok there :)

Answer 12

thank you! are you sure 7/2 or 1/2 isnt included?

Answer 13

and by using this method you know what they may be... a little substitute shows 1 and -1 aren't by 2 is

Answer 14

they should be as they are in the p/q form.....

Answer 15

it could be ± one–half, ± 1, ± 2, ± seven over two, ± 7, ± 14

Answer 16

so using the possible x-7, didn't work btw
so I used x-2 and => \(\bf \begin{matrix}
2 |& 2 & -4 & -7 & 14\\
\ |&& 4 & 0 & -14\\
\hline\\
\ |&2& 0& -7&0
\end{matrix}\)

Answer 17

so there, x-2 is a root
and the remainder is \(\bf 2x^2-7x\)

Answer 18

thats correct..

Answer 19

then you can just use the quadratic for that \(\bf 2x^2-7x+0\)

Answer 20

hmmm wait a second, dohh

Answer 21

omg so what answer is it lol

Answer 22

my x = 0, so it should be \(\bf 2x^2 + 7\)

Answer 23

so you don't need a quadratic formula for that

Answer 24

right off the bat you can tell it'll be 2 complex ones

Answer 25

± one–half, ± 1, ± 2, ± seven over two, ± 7, ± 14?

Answer 26

j you still arent getting it its not asking for complex positive or negative

Answer 27

am I ahead of myself?

Answer 28

its just asking for all the possible zeros

Answer 29

heheh, I get it heheh, yes you're right, yes, those are the answers
they're not asking on the actual roots :)

Answer 30

@jdoe0001 the question just asks what the possible rational roots are.

by using the rational root theorem and listing the possibles... thats all you need
to do.

Answer 31

and yes, as campbell_st said, you are correct, those are it

Answer 32

so its for sure ± one–half, ± 1, ± 2, ± seven over two, ± 7, ± 14

Answer 33

first I misread as find the positive roots
then I misread as find the roots
lol

Answer 34

yes just write all the arrangements of p/q

remember a rational number is any number that can be written as a fraction.