Question

2x^6+8x^2=9 rational roots

Answer 1

\[\pm1 \pm3 \pm9\]

Answer 2

\[\pm1 \pm2 \pm4 \pm 8\]

Answer 3

\[\pm1 \pm3 \pm9 \pm1/2 \pm3/2\pm9/2\]

Answer 4

\[\pm1 \pm2 \pm4 \pm8 \pm1/2 \pm1/4 \pm1/8\]

Answer 5

@jigglypuff314

Answer 6

@jim_thompson5910

Answer 7

@Nnesha

Answer 8

@jdoe0001

Answer 9

got to find the constant term

Answer 10

and leading coefficient

Answer 11

form p/q

Answer 12

ohhh hmmm

Answer 13

right.... so.... hmm ok... so that should be simple
just all the possible combinations?

Answer 14

yea

Answer 15

rational root theorem

Answer 16

http://www.mathwords.com/r/rational_root_theorem.htm

Answer 17

\(\large {
2x^6+8x^2=9 \implies
\begin{array}{llll}
2x^6+8x^2-&9=0\\
\uparrow &\uparrow \\
p&q
\end{array}
}\)

Answer 18

so just get the factors for each
set the factors for "p" over the factors "q"

keep in mind that "1" is also a factor

Answer 19

1 3 9

Answer 20

actaully

Answer 21

?

Answer 22

\(\large {
2x^6+8x^2=9 \implies
\begin{array}{llll}
2x^6+8x^2-&9=0\\
\uparrow &\uparrow \\
q&p
\end{array}
}\)

Answer 23

only way I see it as 1 3 9

Answer 24

so "p factors are 3, 3, 1 and 9
and "q" factors are 2, 1 only

so p/q factors will be \(\bf \pm\cfrac{3,3,1,9}{2,1}\)

Answer 25

\(\large {
\pm\cfrac{3,3,1,9}{2,1}\implies
\begin{cases}
\pm \frac{3}{2}\\
\pm \frac{3}{1}\\
\pm \frac{1}{2}\\
\pm \frac{1}{1}\\
\pm \frac{9}{2}\\
\pm \frac{9}{1}\\
\end{cases}
}\)

Answer 26

so it is C

Answer 27

so.. .that'd be p/q

Answer 28

c seems like a reasonable answer

Answer 29

dunno what C is, but anyhow... that's the p/q

if C shows that, then that's the one

keep in mind that the choices are just guides
they should match your answer, not the other way around