Need help with basic factoring equation

Question

I'm just trying to brush up on my factoring of quadratic equations.

$$\frac1{x+3} + \frac1{x^2 + 5x +6}$$

$$\frac1{x+3} + \frac1{(x+2)(x+3)}$$

$$\frac{(x+2)(x+3) + (x+3)}{(x+2)(x+2)(x+3)}$$

Then I'm stuck. I think you can factor out the $(x+2)(x+3)$ from top and bottom?

Answer 1

Between your second and third step it can be improved: since you have $ \frac1{x+3} + \frac1{(x+2)(x+3)} $ in the second step, you want to give all fractions a common denominator - the least common denominator. Since both terms have a $x+3$ in the denom. already, multiply the first term by $\frac{x+2}{x+2}$, and add the fractions to get $$ \frac{x+2}{(x+2)(x+3)} + \frac 1{(x+2)(x+3)} = \frac{x+3}{(x+2)(x+3)} = \frac{1}{x+2}. $$

Answer 2

The top can be rewritten as ((x+2)+1)(x+3) = (x+3)^2. The bottom should be (x+2)(x+3)(x+3), if you go with the "redundant" approach. Two (x+3)'s, not two (x+2)'s. Then you can factor (x+3)^2 from top and bottom and are left with 1/(x+2), as others have noted.

Answer 3

$$\begin{align} &{1\over{x+3}}+{1\over{x^2+5x+6}}\to\\ &{1\over{x+3}}+{1\over{(x+2)(x+3)}}\to\\ &{x+2+1\over{(x+2)(x+3)}}\to\\ &{x+3\over{(x+2)(x+3)}}\to\\ &1\over{x+2} \end{align}$$