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Help.

NotSoSmart  Dec 7, 2017
 #1
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We have

 

1d^6   -  6d^5*(5y)  + 15d^4*(5y)^2  -  20d^3*(5y)^3 + 15d^2(5y)^4 - 6d*(5y)^5 + 1(5y)^6

 

d^6  - 30d^5y + 375d^4y^2 - 2500d^3y^3 + 9375d^2y^3 -  18750dy^5 + 15625y^6

 

 

x^4  -  6x^2  - 7x  - 6 = 0 

 

Using the Rational roots Theorem, 3 is a root....

 

Using synthetic division  to find the remaining polynomial, we have

 

 

 

3  [  1   0     -6      - 7     - 6  ]

            3      9        9        6

       __________________

       1   3    3         2        0

 

 

So  we have

 

x^3  +  3x^2  +  3x   +  2       and we can write

 

x^3  +  3x^2  +  2x   + x + 2     factor

 

x ( x^2 + 3x + 2)  +  1 (x + 2)

 

x (x + 2)(x + 1) +  1 ( x + 2)

 

(x + 2)  [ x(x + 1) + 1 ]

 

(x + 2) [ x^2 + x + 1]  

 

So -2 is the other real root

 

And  using the quadratic formula....the roots of the other polynomial are

 

[-1 ± i√3] / 2

 

 

 

cool cool cool

CPhill  Dec 7, 2017

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