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cubic equation omg

z = x+iy this is complex number question

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

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

There will be one + ve real root.)

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

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

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

You get 2z^2 + 2z +3

Sum of the Roots , product of the roots

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

Now we have found one solutions z = 2

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

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

Gud work...@baddinlol

I am going to complete the square

polynomials are fun aren't they

nop...@baddinlol use quadratic formula

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

(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

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

well done baddinol thanks

np!

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

i did but i thought there must be a faster way

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

Are you allowed to use a cas calculator?

no