## anonymous 5 years ago Solve the equation 2(x^4)+3(x^3)-7(x^2)+3x-9=0 by first finding any rational roots. Thanks!

1. anonymous

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

2. anonymous

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

3. anonymous

1 does not work

4. anonymous

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

5. anonymous

sure.

6. anonymous

-1 does not work either

7. anonymous

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

8. anonymous

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

9. anonymous

-3 works

10. anonymous

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

11. anonymous

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

12. anonymous

depressed?

13. anonymous

you will get $p(x)=(x+3)q(x)$

14. anonymous

you find $q(x)$ by dividing

15. anonymous

in fact i have done it.

16. anonymous

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

17. anonymous

both. if you do not mind.

18. 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

19. anonymous

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

20. anonymous

k you were ahead of me

21. 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 $-\frac{3}{2}$

22. anonymous

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

23. anonymous

try 3/2 and you will get no remainder

24. anonymous

yes, i found that.

25. anonymous

when do I stop finding roots?

26. anonymous

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

27. anonymous

or $p(x)=(x+3)(2x-3)(x^2+1)$

28. anonymous

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

29. 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

30. anonymous

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

31. anonymous

yes because the roots of $x^2+1$ are i and -i

32. 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

33. anonymous

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

34. anonymous

oh no

35. anonymous

they have a specific meaning

36. anonymous

square root of -1.

37. anonymous

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

38. anonymous

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

39. anonymous

ahhhhhh. ok. thanks a ton!!

40. anonymous

not a real number

41. anonymous

welcome. hope it was clear

42. anonymous

definitely.

Find more explanations on OpenStudy