How do you factor #y= x^2-3x-1# ?
1 Answer
Dec 24, 2015
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
#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)#
