Professor Leonard, thank you for a well explained and organized lecture/video on Creating Polynomials from Complex Solutions using the well-known complex conjugate pairs. These problems are fun and easy to solve from start to finish.
I pray you're doing well during this pandemic. We're all struggling but I know we'll get through it. From all of us continuing to take classes and continuing our education during all of this, thank you.
Thank you professor, anxiously waiting for the linear algebra 1 course.Currently winding up with the calculus 1 course on your channel (bliss!). You're a legend!
No, Since i^2 =-1 , in order to find i, we need to take square root on both sides. By doing so we will end up with the result i=sqrt.-1 which is not possible under real number system, so mathematicians introduced i for this
When factoring a polynomial, you can break it down to a linear function and an irreducible function for example given a function: 2x^3-x^2+2x-3 you can break it down to its factor term (2x^3-x^2+2x--3)= (x-1)*(2x^2+x+3) for x-1=0, you can see that there is an xint at 1 but for f(x)=2x^2+x+3 when you try to solve for f(x)=0, aka its x int you get a complex number, indicating that the graph of that line does not touch the x axis. hence the term irreducible quadratic because if you try to break it down even more aka factor.... you won't find solutions for it in the real term - i might be oversimplifying but I do feel I am correct here but as a note this is video no 35 out an entire lecture series so you will have to go back and watch the earlier videos to make sense of it all. best!
I'm a little bit confused on how you got x^2-2x+1+1 when you distributed at 37:36. If you could explain or link me another video that explains how you got that I would appreciate it a lot.
