How do you simplify #(6x^2-4x-3)/(3x^2+x)#?

1 Answer
Jan 8, 2017

Answer:

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

Explanation:

Notice that:

#3x^2+x = x(3x+1)#

Then #x# is not a factor of the numerator, but #(3x+1)# might be.

Let's try:

#6x^2-4x-3 = (6x^2+2x)-(6x+2)-1#

#color(white)(6x^2-4x-3) = 2x(3x+1)-2(3x+1)-1#

#color(white)(6x^2-4x-3) = (2x-2)(3x+1)-1#

No. So there is no cancellable factor and we cannot substantially simplify the given rational expression.

About the most we can do is separate out the quotient #2# as follows:

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

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

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