Complex Roots of Polynomials

We use the quadratic formula to find all complex roots of polynomials. This is Chapter 3, Problem 8 of Math 1141 Algebra notes. Presented by Thanom Shaw of the School of Mathematics and Statistics, UNSW.

[@andrewhl]

7 years later and this video still helps so much

[@anaoc5651]

Thank you so, so much!

[@sathvikmalgikar2842]

Great video mam

[@chandrasai1990]

i never felt more happy

[@NORTHEASTGAMING-er9go]

Beautiful teacher help me feel in love on both math and teacher.

[@kenetworu8028]

That's great 🙏

[@shawnisconyakonda5415]

thanx ma'm ub really good

[@trkingdavid4943]

wawu well explained

[@YO-kq6dt]

Thanks

[@user-et5iu6su3w]

10/10 very helpful

[@rexd1624]

ma'am what about 4 terms? can i still use quadratic formula?

[@rexd1624]

@Lachlan Harris thanks.man!

[@duygurechani290]

Can you help me solve this kind of complex equation: x^4+sqr2*x^2+1=0. Thank you

[@abdulrahmantaher2066]

@MathsStatsUNSW I believe that a mistake has been made. The difference of two squares in real numbers is (a-b)(a+b) = a^2 - b^2 , but in the complex plane, (a-bi)(a+bi) = a^2 + b^2 due to the presence of i.

[@ronaldmayland4133]

can you please elaborate?

[@ashvinbalaraman2767]

Her method seems valid to me. She changed the sign from z^2+1 to z^2-i^2 because we know -i^2 is simply +1. From there she deduced z^2-i^2 equals (z+i)*(z-i) from the difference of two squares as i*(-i) resolves to -i^2 and the “iz” terms cancel.

[@AbdulKarim-wl6cd]

Why do u use z in the place of x

[@supersonic174]

use z for complex numbers

[@ananejean-zl8ie]

Very nice

[@almakahmed96]

all the love

[@kyrilcouda]

Lol... everyone keeps using quadratic formula. I need to know more than that!

[@malwandekhumalo9413]

To us writing tmr 😀

[@riakmacholariakkeer916]

your method has not convince me in second question.

[@XERXES_Arta]

Please is i = -1 or i = root of -1

[@Shubham_pandey-nk1un]

i = √-1

[@isi6402]

i^2= -1 and i= √-1