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jpjonesBest ResponseYou've already chosen the best response.0
z = x+iy this is complex number question
 one year ago

baddinlolBest ResponseYou've already chosen the best response.1
Well first of all you can find out that P(2)=0
 one year ago

baddinlolBest ResponseYou've already chosen the best response.1
Therefore z2 must be a factor of 2z^3  2z^2  z  6
 one year ago

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

jpjonesBest ResponseYou've already chosen the best response.0
why is it that you have fed 2 into the equation, is this a random number
 one year ago

baddinlolBest ResponseYou've already chosen the best response.1
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
 one year ago

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

baddinlolBest ResponseYou've already chosen the best response.1
Then when you divide 2z^3 + 2z^2z6 by z2
 one year ago

baddinlolBest ResponseYou've already chosen the best response.1
You get 2z^2 + 2z +3
 one year ago

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

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

baddinlolBest ResponseYou've already chosen the best response.1
Now we have found one solutions z = 2
 one year ago

baddinlolBest ResponseYou've already chosen the best response.1
Other two are given by 2z^2+2z+3=0
 one year ago

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

baddinlolBest ResponseYou've already chosen the best response.1
I am going to complete the square
 one year ago

jpjonesBest ResponseYou've already chosen the best response.0
polynomials are fun aren't they
 one year ago

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

baddinlolBest ResponseYou've already chosen the best response.1
2(z^2 + z + 3/2) = 0 z^2 + z + 3/2 = 0 (z + 1/2)^2 + 5/4 = 0
 one year ago

baddinlolBest 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
 one year ago

jpjonesBest ResponseYou've already chosen the best response.0
that will be =+orsqrt b^24(a)(c)/(2a)
 one year ago

jpjonesBest ResponseYou've already chosen the best response.0
well done baddinol thanks
 one year ago

baddinlolBest ResponseYou've already chosen the best response.1
Did you know about putting in numbers to find p(Z) = 0?
 one year ago

baddinlolBest ResponseYou've already chosen the best response.1
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^32z^2z6
 one year ago

jpjonesBest ResponseYou've already chosen the best response.0
i did but i thought there must be a faster way
 one year ago

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

baddinlolBest ResponseYou've already chosen the best response.1
Are you allowed to use a cas calculator?
 one year ago
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