Is it possible to factor #y= x^2 -x - 11 #? If so, what are the factors?

1 Answer
Dec 22, 2015

Answer:

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 #a=(x-1/2)# and #b = sqrt(45)/2# ...

#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 #a, b >= 0# then #sqrt(ab) = sqrt(a)sqrt(b)#, so we can simplify:

#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)#