Polynomial Equation

Hello Melody,

how did you know that a+b+y+o=7?? 

Consider the trinomal 

$2x^2-7x-4=0\\ \text{this factorises to }\;\;(2x+1)(x-4)=0\\ \text{so the roots }\alpha\;\;and\;\;\beta \;\; are \;\;\frac{-1}{2}\;\;and \;\;+4\\ \alpha+\beta=\frac{-1}{2}+4=3\frac{1}{2}=\frac{7}{2}\\ \alpha*\beta=\frac{-1}{2}*4=-2=-\frac{4}{2}\\ \\~\\ $

Compare thes numbers to the coefficients in the original equation.  

What you can see is not a coincidence.

And it can be extanded to polynomials of ANY degree.

So in your polynomial

$Ax^4+Bx^3+Cx^2+Dx^1+E=0$

If there are 4 real roots and they are

  $\alpha,\beta,\gamma, \delta\;\;\;then\\ \alpha+\beta+\gamma+ \delta=-\frac{B}{A}\\ \text{The sum of the product of root pairs will be}\quad +\frac{C}{A}\\ \text{The sum of the product of root tripples will be}\quad -\frac{D}{A}\\ and\;\;lastly\\ \alpha\beta\gamma \delta=+\frac{E}{A}\\ $

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I am not sure how it works if the roots are not real.  I can't remember thinking about that before.

Here is a video that covers some of this.

https://www.khanacademy.org/math/math-for-fun-and-glory/aime/2003-aime/v/sum-of-polynomial-roots-proof

Coding:

2x^2-7x-4=0\\
\text{this factorises to }\;\;(2x+1)(x-4)=0\\
\text{so the roots }\alpha\;\;and\;\;\beta \;\; are \;\;\frac{-1}{2}\;\;and \;\;+4\\
\alpha+\beta=\frac{-1}{2}+4=3\frac{1}{2}=\frac{7}{2}\\
\alpha*\beta=\frac{-1}{2}*4=-2=-\frac{4}{2}\\

Ax^4+Bx^3+Cx^2+Dx^1+E=0
 

Thank you, that was really useful!! 

You are welcome.  :)