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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
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
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...
I am going to complete the square
polynomials are fun aren't they
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
Did you know about putting in numbers to find p(Z) = 0?
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
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?