How do you factor #x^{2}-x-5#?

1 Answer
George C.
Aug 21, 2017

#x^2-x-5 = 1/4(2x-1-sqrt(21))(2x-1+sqrt(21))#

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 #a=(2x-1)# and #b=sqrt(21)# as follows:

#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))#

Related questions
Impact of this question
1417 views around the world
You can reuse this answer
Creative Commons License