How do you factor #y= x^2-3x-1# ?

1 Answer
Dec 24, 2015

Answer:

It requires irrational coefficients, but...

Complete the square and use the difference of squares identity to find:

#y = x^2-3x-1#

#= (x-3/2-sqrt(13)/2)(x-3/2+sqrt(13)/2)#

Explanation:

Note that:

#(x-3/2)^2 = x^2-3x+9/4#

Also, the difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We use this with #a = (x-3/2)# and #b=sqrt(13)/2# below...

#y = x^2-3x-1#

#= x^2-3x+9/4-9/4-1#

#= (x-3/2)^2-13/4#

#= (x-3/2)^2-(sqrt(13)/2)^2#

#= ((x-3/2)-sqrt(13)/2)((x-3/2)+sqrt(13)/2)#

#= (x-3/2-sqrt(13)/2)(x-3/2+sqrt(13)/2)#