Solve the equation x^5-5x^4-5x^3+25x^2+4x-20=0, given that it's roots are of the form a,-a,b,-b,c...
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If a and -a and b and -b are factors, then x^2-a^2 and x^2-b^2 are factors.

The polynomial can be written (x^2-a^2)(x^2-b^2)(x-c)=x^5-5x^4-5x^3+25x^2+4x-20=0.

(x^4-x^2(a^2+b^2)+a^2b^2)(x-c)=x^5-x^3(a^2+b^2)+a^2b^2x-cx^4+cx^2(a^2+b^2)-ca^2b^2=x^5-5x^4-5x^3+25x^2+4x-20=0.

Equating coefficients:

x^4: -c=-5, so c=5.

x^3: -a^2-b^2=-5, a^2+b^2=5.

x^2: c(a^2+b^2)=25 confirming c=5 and a^2+b^2=5.

x: a^2b^2=4.

Constant: -ca^2b^2=-20, confirming above.

a^2+4/a^2=5; a^4-5a^2+4=0=(a^2-4)(a^2-1)=0 so a=2 or 1 and b=1 or 2.

The roots are -1, -2, 1, 2, 5 (x+1)(x-1)(x+2)(x-2)(x-5)=0.
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