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• so, x^2-5x-8 first solve it x^2-5x-8=0 D= b^2 - 4*a*c = 25-4*1*-8 = 57 your x1 then becomes = (-b+ √D) /(2a) = (5 + √57)/2 x2= (-b- √D) /(2a) = (5- √57)/2 so x^2-5x-8 = (x - (5- √57)/2) ( x - (5 + √57)/2) Hope it helped

  in How do you factor x^2-5x-8? Comment by xwonderer June 2021
• so, x^2-5x-8 first solve it x^2-5x-8=0 D= b^2 - 4*a*c = 25-4*1*-8 = 57 your x1 then becomes = (-b+ √D) /(2a) = (5 + √57)/2 x2= (-b- √D) /(2a) = (5- √57)/2 so x^2-5x-8 = (x - (5- √57)/2) ( x - (5 + √57)/2) Hope it helped

  in How do you factor x^2-5x-8? Comment by xwonderer June 2021
• so, x^2-5x-8 first solve it x^2-5x-8=0 D= b^2 - 4*a*c = 25-4*1*-8 = 57 your x1 then becomes = (-b+ √D) /(2a) = (5 + √57)/2 x2= (-b- √D) /(2a) = (5- √57)/2 so x^2-5x-8 = (x - (5- √57)/2) ( x - (5 + √57)/2) Hope it helped

  in How do you factor x^2-5x-8? Comment by xwonderer May 2021
• so, x^2-5x-8 first solve it x^2-5x-8=0 D= b^2 - 4*a*c = 25-4*1*-8 = 57 your x1 then becomes = (-b+ √D) /(2a) = (5 + √57)/2 x2= (-b- √D) /(2a) = (5- √57)/2 so x^2-5x-8 = (x - (5- √57)/2) ( x - (5 + √57)/2) Hope it helped

  in How do you factor x^2-5x-8? Comment by xwonderer May 2021