# Answer to Question #26165 in Algebra for fatma

Question #26165

Solve the reciprocal equation 6x6-25x5+31x4-31x2+25x-6=0

Expert's answer

We can rewrite the equation

(x-2)(x-1)(x+1)(2x-1)(3x^2 - 5x +3)=0

It's possible if one of the factors equals zero. Therefore we obtain the following equation: x-2=0 or x-1=0or x+1=0 or 2x-1=0 or 3x^2 - 5x +3 =0

So we have 4 real solutions and 2 complex.

Real:

x=2

x=1

x=-1

x=1/2

Complex:

x=1/6(5-i*sqrt(11))

x=1/6(5+i*sqrt(11))

(x-2)(x-1)(x+1)(2x-1)(3x^2 - 5x +3)=0

It's possible if one of the factors equals zero. Therefore we obtain the following equation: x-2=0 or x-1=0or x+1=0 or 2x-1=0 or 3x^2 - 5x +3 =0

So we have 4 real solutions and 2 complex.

Real:

x=2

x=1

x=-1

x=1/2

Complex:

x=1/6(5-i*sqrt(11))

x=1/6(5+i*sqrt(11))

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