Question

Solve the equation 2(x^4)+3(x^3)-7(x^2)+3x-9=0 by first finding any rational roots.

Thanks!

Answer 1

possible rational roots are \[\pm1, \pm3, \pm9, \pm\frac{1}{2}, \pm\frac{3}{2},\pm\frac{9}{2}\]

Answer 2

a lot to check actually so i would start with the easy ones : 1 and -1

Answer 3

1 does not work

Answer 4

if you give me a minute i will find them for you

Answer 5

sure.

Answer 6

-1 does not work either

Answer 7

don't you need to use synthetic division to find the other roots as well?

Answer 8

well you need to find at least one to start. you can use synthetic division sure. i was going to cheat

Answer 9

-3 works

Answer 10

once you have -3 you can use synthetic division to factor

Answer 11

and then what do I do w/ the depressed equation?

Answer 12

depressed?

Answer 13

you will get \[p(x)=(x+3)q(x)\]

Answer 14

you find \[q(x)\] by dividing

Answer 15

in fact i have done it.

Answer 16

would you like the answer or do you want to find it?

Answer 17

both. if you do not mind.

Answer 18

ok. first of all \[p(-3)=0\] you can check this by synthetic division, or by replacing x by -3 which is kind of a pain

Answer 19

ok, i have done the division.

the equation is 2x^3-3x^2+2x-3=0.

Answer 20

k you were ahead of me

Answer 21

now the possible rational roots have narrowed somewhat. we know 1 and -1 don't work. we found -3. 3 does not work. now you can try \[\frac{3}{2}\] and \[-\frac{3}{2}\]

Answer 22

i give you a hint, it is not the negative one!

Answer 23

try 3/2 and you will get no remainder

Answer 24

yes, i found that.

Answer 25

when do I stop finding roots?

Answer 26

ok and you get \[p(x)=(x+3)(x-\frac{3}{2})(2x^2+2)\]

Answer 27

or
\[p(x)=(x+3)(2x-3)(x^2+1)\]

Answer 28

now you can stop because \[x^2+1\] does not factor over the real numbers

Answer 29

if you are using complex numbers then i guess you can write \[p(x)=(x+3)(2x-3)(x+i)(x-i)\] but that is kind of silly because we are working with a real valued polynomial

Answer 30

is it also b/c the original equation is to the third degree? somehow, I also found +i, -i...

Answer 31

yes because the roots of \[x^2+1\] are i and -i

Answer 32

we have found the 4 zeros, 2 real and two complex. of course you can stop now because a polynomial of degree 4 can have a most 4 zeros

Answer 33

ohhhh, so +i and -i can b considered any number?

Answer 34

oh no

Answer 35

they have a specific meaning

Answer 36

square root of -1.

Answer 37

\[x^2+1=0\]
\[x^2=-1\]
\[x=\pm\sqrt{-1}\]

Answer 38

it is just a symbol for \[\sqrt{-1}\]

Answer 39

ahhhhhh. ok. thanks a ton!!

Answer 40

not a real number

Answer 41

welcome. hope it was clear

Answer 42

definitely.