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x5-x4-3x3+3x2-4x+4=0

Stein101  May 22, 2017
edited by Stein101  May 22, 2017
 #1
avatar+92368 
+1

x^5-x^4-3x^3+3x^2-4x+4=0

 

If we can add the coefficients and the result  = 0, then 1 is a root

 

So using synthetic division to find a reduced polynomial, we have

 

 

1  [  1    -1   - 3    +3   -4     4  ]

              1     0    -3     0     -4

     _____________________

      1     0    -3     0     -4    0

 

So.......the remaining polynomial   is       x^4  -  3x^2  -  4     factor this

 

(x^2 - 4) (x^2 + 1)    =

 

(x + 2) ( x - 2)  (x^2  + 1)

 

Setting the first two terms to  0 and solving for x we get that x  = -2  and x  = -2....and these are two other real roots

 

Finally.........set x^2 +  1  to 0

 

x^2  + 1  =  0      subtract 1 from both sides

 

x^2  =  -1        and x  =   i       and x  = -i      are the non-real roots

 

So the zeroes are      -2, 1, 2 , i, -i

 

 

cool cool cool

CPhill  May 22, 2017

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