How do you factor #x^{2}-x-5#?
1 Answer
Aug 21, 2017
Explanation:
We can factor this quadratic by completing the square and using the difference of squares identity:
#a^2-b^2 = (a-b)(a+b)#
with
#4(x^2-x-5) = 4x^2-4x+1-21#
#color(white)(4(x^2-x-5)) = (2x-1)^2-(sqrt(21))^2#
#color(white)(4(x^2-x-5)) = ((2x-1)-sqrt(21))((2x-1)+sqrt(21))#
#color(white)(4(x^2-x-5)) = (2x-1-sqrt(21))(2x-1+sqrt(21))#
So:
#x^2-x-5 = 1/4(2x-1-sqrt(21))(2x-1+sqrt(21))#