Is it possible to factor #y=3x^2 + 9x-6#? If so, what are the factors?

1 Answer
Feb 22, 2017

Answer:

#3x^2+9x-6 = 3(x+3/2-sqrt(17)/2)(x+3/2+sqrt(17)/2)#

Explanation:

Complete the square then use the difference of squares identity:

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

with #a=(2x+3)# and #b=sqrt(17)# as follows:

#3x^2+9x-6 = 3/4(4x^2+12x-8)#

#color(white)(3x^2+9x-6) = 3/4((2x)^2+2(2x)(3)+3^2-17)#

#color(white)(3x^2+9x-6) = 3/4((2x+3)^2-(sqrt(17))^2)#

#color(white)(3x^2+9x-6) = 3/4(2x+3-sqrt(17))(2x+3+sqrt(17))#

#color(white)(3x^2+9x-6) = 3(x+3/2-sqrt(17)/2)(x+3/2+sqrt(17)/2)#