How do you factor the expression #x^2+3x+1#?

1 Answer
Sep 2, 2016

Answer:

#x^2+3x+1 = (x+3/2-sqrt(5)/2)(x+3/2+sqrt(5)/2)#

Explanation:

We can complete the square and use the difference of squares identity, which may be written:

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

with #a=(x+3/2)# and #b=sqrt(5)/2# as follows:

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

#color(white)(x^2+3x+1) = (x+3/2)^2-(sqrt(5)/2)^2#

#color(white)(x^2+3x+1) = ((x+3/2)-sqrt(5)/2)((x+3/2)+sqrt(5)/2)#

#color(white)(x^2+3x+1) = (x+3/2-sqrt(5)/2)(x+3/2+sqrt(5)/2)#