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jpjones
 3 years ago
find all roots of the polynomial
p(z) = 2z^32z^2z6
jpjones
 3 years ago
find all roots of the polynomial p(z) = 2z^32z^2z6

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jpjones
 3 years ago
Best ResponseYou've already chosen the best response.0z = x+iy this is complex number question

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.1Well first of all you can find out that P(2)=0

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.1Therefore z2 must be a factor of 2z^3  2z^2  z  6

Yahoo!
 3 years ago
Best ResponseYou've already chosen the best response.0There will be one + ve real root.)

jpjones
 3 years ago
Best ResponseYou've already chosen the best response.0why is it that you have fed 2 into the equation, is this a random number

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.1When 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

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.1So if P(2) gives you zero, z2 must be a factor of p(z)

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.1Then when you divide 2z^3 + 2z^2z6 by z2

Yahoo!
 3 years ago
Best ResponseYou've already chosen the best response.0Sum of the Roots , product of the roots

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.1So we have the following expression (2z^2 +2z +3)(z2)=0

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.1Now we have found one solutions z = 2

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.1Other two are given by 2z^2+2z+3=0

Yahoo!
 3 years ago
Best ResponseYou've already chosen the best response.0and the other sol^n will be Complex numbers...

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.1I am going to complete the square

jpjones
 3 years ago
Best ResponseYou've already chosen the best response.0polynomials are fun aren't they

Yahoo!
 3 years ago
Best ResponseYou've already chosen the best response.0nop...@baddinlol use quadratic formula

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.12(z^2 + z + 3/2) = 0 z^2 + z + 3/2 = 0 (z + 1/2)^2 + 5/4 = 0

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.1(z + 1/2)^2 = 5/4 z + 1/2 = sqrt(5)i /2, sqrt(5)i /2 z = (sqrt(5)i  1)/2, (1sqrt(5)i)/2

jpjones
 3 years ago
Best ResponseYou've already chosen the best response.0that will be =+orsqrt b^24(a)(c)/(2a)

jpjones
 3 years ago
Best ResponseYou've already chosen the best response.0well done baddinol thanks

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.1Did you know about putting in numbers to find p(Z) = 0?

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.1By 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^32z^2z6

jpjones
 3 years ago
Best ResponseYou've already chosen the best response.0i did but i thought there must be a faster way

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.1SO i tried 1, 1, 2 and when i got upto 2 i stopped because i found P(2)=0

baddinlol
 3 years ago
Best ResponseYou've already chosen the best response.1Are you allowed to use a cas calculator?
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