How do you factor #x^2+9x+3#?

1 Answer
George C.
Sep 14, 2016

#x^2+9x+3 = (x+9/2-sqrt(69)/2)(x+9/2+sqrt(69)/2)#

Explanation:

Complete the square then use the difference of squares identity:

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

with #a=(x+9/2)# and #b=sqrt(69)/2# as follows:

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

#color(white)(x^2+9x+3) = (x+9/2)^2-81/4+12/4#

#color(white)(x^2+9x+3) = (x+9/2)^2-69/4#

#color(white)(x^2+9x+3) = (x+9/2)^2-(sqrt(69)/2)^2#

#color(white)(x^2+9x+3) = ((x+9/2)-sqrt(69)/2)((x+9/2)+sqrt(69)/2)#

#color(white)(x^2+9x+3) = (x+9/2-sqrt(69)/2)(x+9/2+sqrt(69)/2)#

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