How do you factor #2x^2-5x+1#?

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Answer 1 · 2017-03-02T08:18:53

#2 x^2 - 5 x + 1 = 2(x - (5 + sqrt 17)/4)(x - (5- sqrt 17)/4)#

#2 x^2 - 5 x + 1 = 2(x^2 - 5/2 x + 1/2)#

Consider #x^2 - 5/2 x + 1/2 = 0# and solve by completing a square
#x^2 - 5/2 x + 1/2 = 0#

#(x -5/4)^2 -(5/4)^2 + 1/2 = 0#

#(x -5/4)^2 -25/16 + 1/2 = 0#

#(x -5/4)^2 -17/16 = 0#

#(x -5/4)^2 = 17/16#

#(x -5/4) = +-sqrt (17/16) = +-sqrt 17/4#

#x = 5/4 +-sqrt 17/4 = (5+- sqrt 17)/4#

#(x^2 - 5/2 x + 1/2) = (x - (5 + sqrt 17)/4)(x - (5- sqrt 17)/4)#

therefore

#2 x^2 - 5 x + 1 = 2(x - (5 + sqrt 17)/4)(x - (5- sqrt 17)/4)#