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2015-05-30T21:31:54+05:30

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Factor theorem says that  if  x - a is a factor of   f(x)  if and only if  f(a) = 0...

 F(x) = x^n  + a^n
  F(-a) = -(1)^n a^n + a^n   =   0   for   odd   n...  as  -1)^n  = -1..
hence  x + a is a factor of  F(x)

 ===================
F(x) =  x⁴  - 4 / x⁴  =  (x⁸ - 4) / x⁴
  Let    a = √√2  = (2)¹/⁴ 
             then  a² = √2  and  a⁴ = 2      then  a⁸ = 4
  
F(a) =  (a⁸ - 4) / a⁴  = 0          =>  x - a  is a factor of  F(x)
F(x)  =  (x - a) (x⁷ + b x⁶ + c x⁵ + d x⁴ + e x³ + f x² + g x + 4/a) / x⁴
        =  [x⁸ + (b-a) x⁷ + (c-ab) x⁶ + (d-ac) x⁵ + (e-ad) x⁴ + (f - ae) x³
                       + (g - af) x² + (4/a - ag) x - 4 ] / x⁴
         = (x⁸ - 4) / x⁴

 => b = a
      c = ab = a² = √2
       d = ac = a³ = 2³/⁴
       e = ad =a⁴ = 2
       f = ae =  a⁵ = 2⁵/⁴
       g =   a⁶ = 2³/²
      
   4/a = ag = a⁷      =>  a⁸ = 4.. which i s  true,,

F(x) =  (x - a) (x⁷ + a x⁶ + √2 x⁵ + a³ x⁴ + 2 x³ + 2⁵/⁴ x² + 2³/² x + 4/a) / x⁴
       =  (x - a) (x³ + a x² + √2 x + 2³/⁴ + 2 x⁻¹ + 2⁵/⁴ /x⁻² + 2√2 x⁻³ + 4/a x⁻⁴)

Factorizing : 
  F(x) = (x⁸ - 4) / x⁴  = (x⁴ - 2) (1 + 2 x⁻⁴) 
         = (x² - √2) (x²  + √2) (1 + 2 x⁻⁴)
         =  (x - a) (x + a) (x² + √2) (a + 2 x⁻⁴)

         where a² = √2
=================================

x+2  is a factor of   F(x) = (x + 1)⁷ + (2x + k)³
  =>  F(-2) = 0   
  =>  -1⁷ + (k-4)³  = 0
  =>  k-4 = 1
  =>  k = 5

===================

divisor    x^2 + x - 12 = (x + 4) (x - 3)
Let    x^3 - 6 x^2 - 15 x + 80
     = (x + 4) (x - 3) *  (x + a)    +    [  b (x + 4)  +  c (x - 3) ]
Thus the remainder after division by divisor is  (b + c) x + 4 b - 3 c

Let F(x) =  x^3 - 6 x^2 - 15 x + 80 - (b + c) x - 4 b +  3 c
   F(-4) = 0  as it should be divisible by x + 4
     =>  -64 - 96 + 60 +  4 (b+c) - 4 b + 3c = 0
     =>    c =  100/7

F(3) = 0    as F(x) is divisible by  x - 3
    =>  27 - 54 - 45 + 80 - 3(b+c) - 4 b + 3 c  =  0
    =>   b =  8/7

So   4 * [ 27 x - 67] /7    must be subtracted from the iven polynomial for the divisor to exatly divide it....


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