## evy15 Group Title Find all the zeros of the eq. 2x^4-5x^3+53x^2-125x+75=0 one year ago one year ago

1. mathmate Group Title

This one can be factorized, so you will need to use the following tools: Descartes rule of signs, factorization theorem, fundamental theorem of algebra Are any or some of these famililar to you? Also, this may indicate that you or your teacher have a typo in the other problem. Please re-check for typos on the other problem.

2. mathmate Group Title

To make things simpler, when the sum of coefficients = 0, it means that (x-1) is a factor.

3. evy15 Group Title

no I dnt know those methods

4. mathmate Group Title

Descartes rule of signs tells us that we have either 4 real positive roots, or two real positive roots and two complex roots, or 4 complex roots.

5. evy15 Group Title

ok

6. mathmate Group Title

http://en.wikipedia.org/wiki/Descartes'_rule_of_signs http://en.wikipedia.org/wiki/Factor_theorem http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra Since you have been missing school for at least one day, it may well be advisable to catch up with some of these topics. Hope you got notes from your friends though.

7. mathmate Group Title

So is (x-1) a factor to the equation?

8. evy15 Group Title

yes

9. mathmate Group Title

So you can do a synthetic division, or a long division to reduce the quartic to a cubic, i.e. find (2*x^4-5*x^3+53*x^2+(-125)*x+75)/(x-1)=?

10. evy15 Group Title

2x^3-3x^2+50x-75

11. mathmate Group Title

Good, now we can rule out the third case (4 complex roots).

12. evy15 Group Title

ok

13. mathmate Group Title

Using the factor theorem, you will need to try factors of the form (ax+b) where a=1 or 2 (from the leading coefficient 2), and b=+/- all possible factors of 75, which are 1,3,5,15,25,75, i.e. 2*2*6=24 possibilities. Once you find that, you can again do the synthetic division or long division to reduce the cubic to a quadratic.

14. evy15 Group Title

what do I divide it by

15. mathmate Group Title

When you find another root (i.e. another factor) of the cubic.

16. evy15 Group Title

oh ok, can I use the calculator for that

17. KingGeorge Group Title

Sorry to bust in here, but $$2x^3-3x^2+50x-75$$ is relatively easily factored by grouping, and thus we can skip the whole guess and check portion of finding roots.

18. mathmate Group Title

In fact, Descartes rule of sign says we don't have to try factors of the form (a+b), only those of the form (a-b) because all real roots are positive.

19. mathmate Group Title

So that reduces to 12 possibilities only.

20. mathmate Group Title

Very true, thanks KingGeorge!

21. KingGeorge Group Title

You group like so $2x^3-3x^2+50x-75=x^2(2x-3)+25(2x-3)$Can evy15 finish the factoring from here?

22. mathmate Group Title

@evy15 are you able to complete the solution from here?

23. evy15 Group Title

can you guide me to what to do next

24. evy15 Group Title

ok hold on

25. evy15 Group Title

x=i+-5 and 3/2

26. evy15 Group Title

i mean =-5i

27. mathmate Group Title

Good job, (+\- 5i). Now you have all 4 zeroes!

28. mathmate Group Title

Can you enumerate the four zeroes?

29. evy15 Group Title

so its 1,+/-5i, and 3/2?

30. evy15 Group Title

or-1

31. mathmate Group Title

Excellent! There you are! Not -1. Your list is good.

32. mathmate Group Title

Big thanks to KingGeorge who cut the work in half!

33. evy15 Group Title

yes thank you guys so much

34. mathmate Group Title

yw! :)

35. KingGeorge Group Title

As an addendum, it's possible to factor the original function by grouping if you're clever. \begin{aligned} 2x^4-5x^3+53x^2-125x+75&=2x^4-5x^3+3x^2+50x^2-125x+75 \\ &=x^2(2x^2-5x+3)+25(2x^2-5x+3) \\ &=(x^2+25)(2x^2-5x+3)\\ &=(x^2+25)(x-1)(2x-3) \end{aligned}

36. KingGeorge Group Title

But this is not an obvious, nor the easiest, way to do it.

37. mathmate Group Title

Yep, I will look more carefully for these cases in the future.

38. evy15 Group Title

can i do the same with x^5-3x^4-24x^3-72x^2+12x=12

39. evy15 Group Title

sorry wrote the wrong one

40. evy15 Group Title

x^5-3x^4-24x^3-72x^2-25x+75=0

41. KingGeorge Group Title

That can definitely be factored by grouping. However, instead of having two groups like I did before, we'll have three groups.

42. KingGeorge Group Title

Your first group will be $x^4(x-3).$Knowing this, can you give me the rest of the factorization?

43. evy15 Group Title

x^4(x-3)-24x^2(x-3)-25(x-3)

44. KingGeorge Group Title

Perfect. And can you finish the rest of the factorization?

45. evy15 Group Title

would it be (x^4-24x^2-25) and (x-3)?

46. KingGeorge Group Title

Right again. Now you just need to factor $$x^4-24x^2-25$$. You can do this by pretending it's a quadratic equation, and not a quartic. So how does $$x^2-24x-25$$ factor?

47. evy15 Group Title

x-25 and x+1

48. KingGeorge Group Title

Perfect. So that means $$x^4-24x^2-25=(x^2-25)(x^2+1)$$. Hence, $x^5-3x^4-24x^3-72x^2-25x+75=(x^2-25)(x^2+1)(x-3)$From here, you just have some sums/differences of squares, and should be able to figure out the roots. I've got to go now, but you seem to be catching the hang of this!