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anonymous
 5 years ago
Solve the equation 2(x^4)+3(x^3)7(x^2)+3x9=0 by first finding any rational roots.
Thanks!
anonymous
 5 years ago
Solve the equation 2(x^4)+3(x^3)7(x^2)+3x9=0 by first finding any rational roots. Thanks!

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0possible rational roots are \[\pm1, \pm3, \pm9, \pm\frac{1}{2}, \pm\frac{3}{2},\pm\frac{9}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a lot to check actually so i would start with the easy ones : 1 and 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you give me a minute i will find them for you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 does not work either

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't you need to use synthetic division to find the other roots as well?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well you need to find at least one to start. you can use synthetic division sure. i was going to cheat

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0once you have 3 you can use synthetic division to factor

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then what do I do w/ the depressed equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you will get \[p(x)=(x+3)q(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you find \[q(x)\] by dividing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in fact i have done it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would you like the answer or do you want to find it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0both. if you do not mind.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok. 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok, i have done the division. the equation is 2x^33x^2+2x3=0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k you were ahead of me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now 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
 5 years ago
Best ResponseYou've already chosen the best response.0i give you a hint, it is not the negative one!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0try 3/2 and you will get no remainder

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when do I stop finding roots?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok and you get \[p(x)=(x+3)(x\frac{3}{2})(2x^2+2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or \[p(x)=(x+3)(2x3)(x^2+1)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now you can stop because \[x^2+1\] does not factor over the real numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you are using complex numbers then i guess you can write \[p(x)=(x+3)(2x3)(x+i)(xi)\] but that is kind of silly because we are working with a real valued polynomial

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it also b/c the original equation is to the third degree? somehow, I also found +i, i...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes because the roots of \[x^2+1\] are i and i

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we 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
 5 years ago
Best ResponseYou've already chosen the best response.0ohhhh, so +i and i can b considered any number?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they have a specific meaning

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x^2+1=0\] \[x^2=1\] \[x=\pm\sqrt{1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is just a symbol for \[\sqrt{1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahhhhhh. ok. thanks a ton!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0welcome. hope it was clear
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