anonymous
  • anonymous
Solve the equation 2(x^4)+3(x^3)-7(x^2)+3x-9=0 by first finding any rational roots. Thanks!
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
possible rational roots are \[\pm1, \pm3, \pm9, \pm\frac{1}{2}, \pm\frac{3}{2},\pm\frac{9}{2}\]
anonymous
  • anonymous
a lot to check actually so i would start with the easy ones : 1 and -1
anonymous
  • anonymous
1 does not work

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anonymous
  • anonymous
if you give me a minute i will find them for you
anonymous
  • anonymous
sure.
anonymous
  • anonymous
-1 does not work either
anonymous
  • anonymous
don't you need to use synthetic division to find the other roots as well?
anonymous
  • anonymous
well you need to find at least one to start. you can use synthetic division sure. i was going to cheat
anonymous
  • anonymous
-3 works
anonymous
  • anonymous
once you have -3 you can use synthetic division to factor
anonymous
  • anonymous
and then what do I do w/ the depressed equation?
anonymous
  • anonymous
depressed?
anonymous
  • anonymous
you will get \[p(x)=(x+3)q(x)\]
anonymous
  • anonymous
you find \[q(x)\] by dividing
anonymous
  • anonymous
in fact i have done it.
anonymous
  • anonymous
would you like the answer or do you want to find it?
anonymous
  • anonymous
both. if you do not mind.
anonymous
  • 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
  • anonymous
ok, i have done the division. the equation is 2x^3-3x^2+2x-3=0.
anonymous
  • anonymous
k you were ahead of me
anonymous
  • 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
  • anonymous
i give you a hint, it is not the negative one!
anonymous
  • anonymous
try 3/2 and you will get no remainder
anonymous
  • anonymous
yes, i found that.
anonymous
  • anonymous
when do I stop finding roots?
anonymous
  • anonymous
ok and you get \[p(x)=(x+3)(x-\frac{3}{2})(2x^2+2)\]
anonymous
  • anonymous
or \[p(x)=(x+3)(2x-3)(x^2+1)\]
anonymous
  • anonymous
now you can stop because \[x^2+1\] does not factor over the real numbers
anonymous
  • 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
  • anonymous
is it also b/c the original equation is to the third degree? somehow, I also found +i, -i...
anonymous
  • anonymous
yes because the roots of \[x^2+1\] are i and -i
anonymous
  • 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
  • anonymous
ohhhh, so +i and -i can b considered any number?
anonymous
  • anonymous
oh no
anonymous
  • anonymous
they have a specific meaning
anonymous
  • anonymous
square root of -1.
anonymous
  • anonymous
\[x^2+1=0\] \[x^2=-1\] \[x=\pm\sqrt{-1}\]
anonymous
  • anonymous
it is just a symbol for \[\sqrt{-1}\]
anonymous
  • anonymous
ahhhhhh. ok. thanks a ton!!
anonymous
  • anonymous
not a real number
anonymous
  • anonymous
welcome. hope it was clear
anonymous
  • anonymous
definitely.

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