Is it possible to factor #y= x^2 -x - 11 #? If so, what are the factors?
1 Answer
Dec 22, 2015
Yes, with irrational coefficients:
#x^2-x-11 = (x-1/2-(3sqrt(5))/2)(x-1/2+(3sqrt(5))/2)#
Explanation:
Complete the square and use the difference of squares identity:
#a^2-b^2 = (a-b)(a+b)#
with
#x^2-x-11#
#=x^2-x+1/4-45/4#
#=(x-1/2)^2-45/4#
#=(x-1/2)^2-(sqrt(45)/2)^2#
#= ((x-1/2)-sqrt(45)/2)((x-1/2)+sqrt(45)/2)#
#= (x-1/2-sqrt(45)/2)(x-1/2+sqrt(45)/2)#
Finally note that if
#sqrt(45) = sqrt(9*5) = sqrt(9)*sqrt(5) = 3sqrt(5)#
So we can write:
#x^2-x-11 = (x-1/2-(3sqrt(5))/2)(x-1/2+(3sqrt(5))/2)#