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Lesson 7 - Firstly what is Vietas formula The idea is to be...

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Firstly, what is Vieta’s formula? The idea is to be able to expand a any polynomial of the form: ( x - r 1 )( x - r 2 ) · · · ( x - r n ) Where n is the degree of the polynomial or the number of roots. To do this, he used something called Symmetric Sums. S 0 = 1 S 1 = r 1 + r 2 + · · · + r n For j < k S 2 = r 1 r 2 + r 1 r 3 + · · · + r j r k + · · · + r n - 1 r n · · · S n = r 1 r 2 · · · r n This gives : ( x - r 1 )( x - r 2 ) · · · ( x - r n ) = S 0 x n - S 1 x n - 1 + S 2 x n - 2 · · · + ( - 1) n S n Enough general theory, here is an example: ( x - 1)( x - 2)( x + 1) S 0 = 1 S 1 = r 1 + r 2 + r 3 = 1 + 2 - 1 = 2 S 2 = r 1 r 2 + r 1 r 3 + r 2 r 3 = 1 * 2 + 1 * ( - 1) + 2 * ( - 1) = - 1 S 3 = r 1 r 2 r 3 = 1 * 2 * ( - 1) = - 6 This gives : ( x - 1)( x - 2)( x + 1) = S 0 x 3 - S 1 x 2 + S 2 x - S 3 = 1 x 3 - 2 x 2 + ( - 1) x - ( - 6) = x 3 - 2 x 2 - x + 6 Of course, this idea goes backwards: If we know r 1 + r 2 + r 3 = 3 , r 1 r 2 + r 1 r 3 + r 2 r 3 = 3 , r 1 r 2 r 3 = 1 Then r 1 , r 2 , r 3 are the three roots of the polynomial x 3 - 3 x 2 + 3 x - 1.
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Factoring specific polynomials There are a few special cases of polynomials where it is easy to find one root.(Includes difference of squares) a n - b n = ( a - b )( a n - 1 + a n - 2 b + · · · ab n - 2 + b n - 1 ) For n odd, this also holds: a n + b n = ( a + b )( a n - 1 - a n - 2 b + · · · - ab n - 2 + b n - 1 ) Assignment 7: 1. What are the solutions to this set of equations:
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