How to Solve Cubic Equations with Complex Roots

Learn the steps required to solve a cubic equation which has one real and two complex roots.

Begin solving our cubic equation by applying the rational roots theorem, this will find us a rational root (if any exist) which we can then use to factor out of our cubic equation through long division. This will reduce our cubic problem to a quadratic which has the same roots as our original cubic equation and give us our remaining two complex roots when solved.

Music by Adrian von Ziegler

[DMlS]

Ok, this does work. However, I think that it is a lot of work for 1 question. What I did was find a factor of 3 that x can equal(try both plus and minus of 1 and 3) to make the left side equal to 0 (as they did) But with the second step, I find it easier to used inspection as apposed to long division. My thought process for this was "What does x(from the x+1) have to be multiplied by to give the first term of the cubic(x^3)? x^2" I then did the same for the last term, "what does 1 in the linear factor have to be multiplied by to give you the 3 in the cubic? 3" I then did the same for the second term of the quadratic, however, this takes a bit more thought. I said "the 1 from linear and x^2 from the quadratic make x^2, what must be added to this to make 3x^2(from the cubic)? an extra 2x^2 is needed. So by multiplying the x by 2x you get the 2x^2 and the quadratic factor is then solved. I then used the quadratic formula(as done in the video) to get the 2 imaginary factors.

To sum it all up, I did basically all that was done in the video, however, instead of doing long division, I used inspection, which in this case seemed a lot easier.

[sharadpatel]

1.574D3 - 4.878D2 - 29.268D - 58.536=0..can u give the method for this kind of problem?

[tbrwy]

does anyone know, if this were a differential equation, what is the format of the general solution? does it induced sin's and cos's?

[pushpendrachaudhary]

Nice, very clearly explained. Thanx

[meghanautiyal]

Thanks but is there any other simple method for finding imaginary roots? Or what happens if all the three roots are imaginary

[mukulbinjola]

i am not getting it for 2x^3+x^2-2x+1 plz help

[XJWill]

A cubic equation may have either all real roots (three or fewer roots that are all real), or 1 real root and two complex conjugate roots. However, rational roots are just a subset of real roots. So some cubic equations that have a real root may not have any rational roots, in which case this method fails.

For example:

x^3 + 4 x^2 + 5 x - 3 = 0

This cubic has one real root and two complex conjugate roots, but the real root is irrational. The exact solution is too long to type here, but the approximate solutions are:

x1 = 0.43343
x2 = -2.2167 + 1.417 i
x3 = -2.2167 - 1.417 i

You'll never find a rational root to that cubic.

[hurshnarayan]

Make a video factorizing a CUBIC polynomial where the constant is a large number ( for e.g., 3000)

OR SIMPLY DO A VIDEO SOLVING THIS EQUATION:

4𝑥^3+100𝑥^2−1214𝑥+3000

[venkatavaishnavreddy]

how to find roots for 4x^3-3x^2+5x-1=0

[hritikpatel]

4s^3+30s^2+34s+78 find roots
If u can

[warrenl]

This is essentially quadratic and not helpful at all. Maybe you should consider change to another title?