Solve the reciprocal equation 6x^(6)-25x^(5)+31x^(4)-31x^(2)+25x-6=0

Question

It must be solved reciprocaly for engineers

Answer 1

6x⁶-25x⁵+31x⁴-31x²+25x-6=0,

6(x⁶-1)-25x(x⁴-1)+31x²(x²-1)=0,

6(x²-1)(x⁴+x²+1)-25x(x²-1)(x²+1)+31x²(x²-1)=0,

(x²-1)(6x⁴+6x²+6-25x³-25x+31x²)=0,

(x²-1)(6x⁴-25x³+37x²-25x+6)=0=(x-1)(x+1)(2x-1)(x-2)(3x²-5x+3).

Obtained by double synthetic division:

2 | 6 -25  37 -25    6

     6  12 -26  22 | -6

½ | 6 -13  11   -3 |  0

      6    3  -5 |   3

      6 -10   6 |   0 = 2(3x²-5x+3)

The quadratic has no real factors, so the real x solution is:

x=-1, 1, ½, or 2. The reciprocal of x is also -1, 1, ½, 2.

Note that 6/x⁶-25/x⁵+31/x⁴-31/x²+25/x-6=0 has the same solution (reciprocal form).