Math, asked by makio, 1 year ago

zeroes of polynomial x cube - x square - x + 2

Answers

Answered by kvnmurty
5
P(x)  = x³ - x² - x + 2  
     let  y = x - 1/3      So x = y + 1/3
P(y) = y³ + y² + y/3 + 1/27 - y² - 1/9 - 2y/3 - y - 1/3 + 2

     P(y) = y³ - 3p y + q = 0 ,          where  p = 4/9  and  q = 43/27

Let    y = s + p/s = (s² + p)/s

   s³ * P(s) = (s⁶ + 3 s⁴p + 3 s² p² + p³) - (3 p s⁴ + 3 p² s²) + s³ q  = 0
                 =>    s⁶ + s³ q + p³ = 0
         s³ = [ - 43/27 +- √(43²/27^2 - 4*4³/9³) ] / 2
         s =  -1.155278    or,   - 0.37007

     x  = y + 1/3  = s + p/s +1/3 =  -1.205556

   P(x) = (x+1.205556) (x² - 2.205556 x + 2/1.205556)

   The other roots are : imaginary as discriminant < 0.

roots:  one negative.   -1.205556
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