How do you factor #x^2 +1 - x#? Algebra Polynomials and Factoring Factoring Completely 1 Answer George C. Jun 7, 2015 #x^2+1-x = x^2-x+1# is of the form #ax^2+bx+c# with #a=1#, #b=-1# and #c=1# The discriminant #Delta# is given by the formula: #Delta = b^2-4ac = (-1)^2-(4xx1xx4) = 1-4 = -3# Since #Delta < 0# the equation #x^2-x+1 = 0# has no real roots and the quadratic #x^2-x+1# has no linear factors with real coefficients. Answer link Related questions What is Factoring Completely? How do you know when you have completely factored a polynomial? Which methods of factoring do you use to factor completely? How do you factor completely #2x^2-8#? Which method do you use to factor #3x(x-1)+4(x-1) #? What are the factors of #12x^3+12x^2+3x#? How do you find the two numbers by using the factoring method, if one number is seven more than... How do you factor #12c^2-75# completely? How do you factor #x^6-26x^3-27#? How do you factor #100x^2+180x+81#? See all questions in Factoring Completely Impact of this question 1139 views around the world You can reuse this answer Creative Commons License