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# Sum of polynomial roots (proof)

Sum of Polynomial Roots. Created by Sal Khan.

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• Why, at , does (x-r1)(x-r2) = 0?
• A polynomial of degree N can be factored into N factors of the form (x-Rn). These factors multiply to get the original polynomial. Since the polynomial is given as equal to 0, you can see how if any of the factors is 0 then the entire product will be zero. So the entire product will be zero and the equation given will be satisfied if X equals any of R1, R2, ... Rn. That's why these are "roots". These are the values of X where the polynomial evaluates to 0. If you think of the function Y = f(x) where f(x) is the polynomials, then graphically, the roots (R1, R2, ..., Rn), are the X values where the function evaluates to 0 or crosses the X axis.
• At , why did Sal divide each side by A? Can anyone explain?
• to get the leading coefficient to be 1. it doesn't change the polynomial at all, but puts it in a nicer format for us.
• Does anyone know how to prove this so-called "Chiu's Method" here:
http://www2.sd38.bc.ca/~fharwood/FactoringNotes
It works, but I don't know how...
• What he does is like using a new variable. 4x^2+5x-6, multiply the polynomial with the leading coefficent '4' and you get: 4*4x^2+4*5x-4*6=(4x)^2+5*(4x)-24. If you think of z=4x you have z^2+5z-24, which factorizes as (z-8)(z+3), and then replace the z with the original 4x: (4x-8)(4x+3). Since you multiplied the polynomial with 4, you have to divide it now by 4, so you get to the original polynomial and you get: (x-2)(4x+3). Dividing only by 4 the first bracket
• x(x^2 + 4) (x^2 - x - 6) = 0 has how many real roots?
• x(x^2 + 4) (x^2 - x - 6) = 0 this means that either x = 0, (x^2 + 4) = 0 or (x^2 - x - 6) = 0
The first: X=0 is a real root so that´s one.
The second: (x^2 + 4)= 0 gives x^2 = -4 which gives x = +/- sqrt(-4) which gives two imaginary roots
The third: (x^2 - x - 6) = 0 which gives x = 0,5 +/-sqrt(6,25) which gives two different real roots.

x(x^2 + 4) (x^2 - x - 6) = 0 has THREE real roots!
• Is in it the "Simon’s Favorite Factoring Trick" formula.
• I'm not sure what you mean, Simon's Favorite Factoring Trick is mainly used for polynomials with multiple variables, such as ab+a+b+1. We can apply it here to get (a+1)(b+1). It's sometimes known as "completing the rectangle".
• If you had to find the sum of the roots of f(x)=x^3-2x+2011, could you be able to find them without the x^2 term?
(1 vote)
• it would be 0, because you have to remember that the x^2 term is still there, but its coefficient is just 0.
(1 vote)
• what is 6x - 5y + 3x + 2y

6x - 5y + 3x + 2y
combine like terms 6x + 3x = 9x and -5y + 2y= -3y
Therefore the sum of the polynomial is 9x - 3y
• when they ask me to let a & b be the solutions of x^2+x-1=0. find a^2+b^2.
how can I solve it?
i know that ab= -1 & a+b=1.. how can i continue
(1 vote)
• a^2+b^2= (a+b)(a+b)-2ab ----( It's true, you can try to expand it, you should remember to apply this when solving this kind of question)
a^2+b^2= (a+b)(a+b)-2ab
=( 1)(1)-2(-1) -----(substitution)
=1+2
=3 **