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## jpjones 3 years ago find all roots of the polynomial p(z) = 2z^3-2z^2-z-6

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1. vikrantg4

cubic equation omg

2. jpjones

z = x+iy this is complex number question

3. baddinlol

Well first of all you can find out that P(2)=0

4. baddinlol

Therefore z-2 must be a factor of 2z^3 - 2z^2 - z - 6

5. Yahoo!

There will be one + ve real root.)

6. jpjones

why is it that you have fed 2 into the equation, is this a random number

7. baddinlol

When you try to factorize polynomials you can try plugging in numbers to see which one of them give you a zero, it is a trial and error way to factorize easily

8. baddinlol

So if P(2) gives you zero, z-2 must be a factor of p(z)

9. baddinlol

Then when you divide 2z^3 + 2z^2-z-6 by z-2

10. baddinlol

You get 2z^2 + 2z +3

11. Yahoo!

Sum of the Roots , product of the roots

12. baddinlol

So we have the following expression (2z^2 +2z +3)(z-2)=0

13. baddinlol

Now we have found one solutions z = 2

14. baddinlol

Other two are given by 2z^2+2z+3=0

15. Yahoo!

and the other sol^n will be Complex numbers...

16. Yahoo!

Gud work...@baddinlol

17. baddinlol

I am going to complete the square

18. jpjones

polynomials are fun aren't they

19. Yahoo!

nop...@baddinlol use quadratic formula

20. baddinlol

2(z^2 + z + 3/2) = 0 z^2 + z + 3/2 = 0 (z + 1/2)^2 + 5/4 = 0

21. baddinlol

(z + 1/2)^2 = -5/4 z + 1/2 = sqrt(5)i /2, -sqrt(5)i /2 z = (sqrt(5)i - 1)/2, (-1-sqrt(5)i)/2

22. jpjones

that will be =+or-sqrt b^2-4(a)(c)/(2a)

23. jpjones

well done baddinol thanks

24. baddinlol

np!

25. baddinlol

Did you know about putting in numbers to find p(Z) = 0?

26. baddinlol

By the way the numbers you plug in have to be a factor of the constant in the equation, which in your equation is -6::::: 2z^3-2z^2-z-6

27. jpjones

i did but i thought there must be a faster way

28. baddinlol

SO i tried 1, -1, 2 and when i got upto 2 i stopped because i found P(2)=0

29. baddinlol

Are you allowed to use a cas calculator?

30. jpjones

no

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