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• you search the roots of equation given x^2 = X so X^2 -8X +15 =0 you have X =5 and X=3 this yields for x1 = sqrt5 x2 = -sqrt5 x3 = sqrt3 x4 =-sqrt3 and the result is (x-sqrt5)(x+sqrt5)(x-sqrt3)(x+sqrt3) you put -2q^2 in factor …

  in How do you factor x^4-8x^2+15? Comment by maussy June 2021
• you search the roots of equation given x^2 = X so X^2 -8X +15 =0 you have X =5 and X=3 this yields for x1 = sqrt5 x2 = -sqrt5 x3 = sqrt3 x4 =-sqrt3 and the result is (x-sqrt5)(x+sqrt5)(x-sqrt3)(x+sqrt3) you put -2q^2 in factor …

  in How do you factor x^4-8x^2+15? Comment by maussy June 2021
• put cos x =y you find 2y^2 -y =0 y (2y-1) =0 two answers y= 0 y =+0.5 translating in angles y = (2k+1) 90° k = integer or y = 60° + 2kpi

  in 2cos^2(x) - cos(x) = 0? Comment by maussy June 2021
• put cos x =y you find 2y^2 -y =0 y (2y-1) =0 two answers y= 0 y =+0.5 translating in angles y = (2k+1) 90° k = integer or y = 60° + 2kpi

  in 2cos^2(x) - cos(x) = 0? Comment by maussy June 2021
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  in Side effects ? Comment by maussy May 2021
• All what you want to know in my link

  in Side effects ? Comment by maussy May 2021
• put cos x =y you find 2y^2 -y =0 y (2y-1) =0 two answers y= 0 y =+0.5 translating in angles y = (2k+1) 90° k = integer or y = 60° + 2kpi

  in 2cos^2(x) - cos(x) = 0? Comment by maussy May 2021
• put cos x =y you find 2y^2 -y =0 y (2y-1) =0 two answers y= 0 y =+0.5 translating in angles y = (2k+1) 90° k = integer or y = 60° + 2kpi

  in 2cos^2(x) - cos(x) = 0? Comment by maussy May 2021