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

- anonymous

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

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- anonymous

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

- anonymous

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

- anonymous

1 does not work

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## More answers

- anonymous

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

- anonymous

sure.

- anonymous

-1 does not work either

- anonymous

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

- anonymous

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

- anonymous

-3 works

- anonymous

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

- anonymous

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

- anonymous

depressed?

- anonymous

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

- anonymous

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

- anonymous

in fact i have done it.

- anonymous

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

- anonymous

both. if you do not mind.

- anonymous

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

- anonymous

ok, i have done the division.
the equation is 2x^3-3x^2+2x-3=0.

- anonymous

k you were ahead of me

- anonymous

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 \[-\frac{3}{2}\]

- anonymous

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

- anonymous

try 3/2 and you will get no remainder

- anonymous

yes, i found that.

- anonymous

when do I stop finding roots?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

oh no

- anonymous

they have a specific meaning

- anonymous

square root of -1.

- anonymous

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

- anonymous

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

- anonymous

ahhhhhh. ok. thanks a ton!!

- anonymous

not a real number

- anonymous

welcome. hope it was clear

- anonymous

definitely.

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