Question 149905
{{{(5x)/(x+2)+(2)/(x^2+4x+4)}}} Start with the given expression.



{{{(5x)/(x+2)+(2)/((x+2)(x+2))}}} Factor {{{x^2+4x+4}}} to get {{{(x+2)(x+2)}}}



{{{((x+2)/(x+2))((5x)/(x+2))+(2)/((x+2)(x+2))}}} Multiply the first fraction by {{{(x+2)/(x+2)}}}



{{{(5x(x+2))/((x+2)(x+2))+(2)/((x+2)(x+2))}}} Combine the fractions.



{{{(5x^2+10x)/((x+2)(x+2))+(2)/((x+2)(x+2))}}} Distribute.



{{{(5x^2+10x+2)/((x+2)(x+2))}}} Since the denominators are equal, we can combine the numerators over the common denominator.



{{{(5x^2+10x+2)/((x+2)^2)}}} Simplify.




So {{{(5x)/(x+2)+(2)/(x^2+4x+4)}}} simplifies to {{{(5x^2+10x+2)/((x+2)^2)}}}



In other words, {{{(5x)/(x+2)+(2)/(x^2+4x+4)=(5x^2+10x+2)/((x+2)^2)}}} where {{{x<>-2}}}