Find Complex Roots of a Cubic Equation z^3 - 3z^2 + z + 5 = 0

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SOLUTION of the equation is z = -1, 2 + i and 2 - i
The final Factored form will be: (z + 1)( z - 2 +1)(z - 2 -i)
Square root of negative numbers leads to complex numbers. They can be represented on Polar coordinate or Argand diagrams. We will discuss, in details, operations and representation of complex numbers in this playlist.
#ComplexNumbers #ImaginaryNumbers #Radicals #Rationalization #IBSL #ImaginaryParts #globalmathinstitute #anilkumar #APmath #ibslmath #edexcel #gcse
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Thank you! for some reason my teacher didn't go over this even tho we had it on our homework

maggiestack
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Wow, this method is fantastic! Thank you very much.

prkfcey
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Dear sir, you have enviable penmanship.

salvatorecastellitto
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Thank you for this video, it helped so much!

julianaott
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Thank you! Great and easy to understand

jiaxifam
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At first when z=1, the value of z in the polynomial should be 1 not 2 .

ankitchakraborty
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And what if I have z^4+2z^3+7z^2-18z+26=0?? :(

swampy
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the final answer is: (z+1)(z-2+i)(z-2-i)

nahidbarghi
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Please help me calculate this
Given that (√3-i) is a square root of the equation Z^9+16(1+i)z^3+a+ib=0
What is the value of a and b?

gatlatwal
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Are these two equations solvable ? z³ + 3z = 1 and z⁵ + z³ = 1 ??

ingeniumpregium
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how to solve these


x^3 - 3x^2 - x - 3


x^3 - x^2 + 2x + 6

amartyabose
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Please please help me . Solve the equation: Z ^2+4Z+20+¡Z(A+1)=0 where A is a constant, has a complex conjugate root. If one of the roots of this quadratic is Z=B+1, where B is areal constant, find the possible values of A

girmatube
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This is the rational roots theorem and this only works if you have a real root that is rational. You could have a real but irrational root and you would never find it with this method.

davebacknolaliki
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The last part has a mistake. Please see it

choti
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If x^4-4x^3+8x^2-8x+4=0
What will be the solution. ..

prasadnijai
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(s^3) +5(s^2)+8s+3 how to get roots of this equation ????

pratikrahate