Precalc factoring

Question

Factor $$x(x+3)^{-\frac{3}{5}} + (x+3)^{\frac{2}{5}}$$

The solution is given as $$(2x+3)(x+3)^{-\frac{3}{5}}$$

I am not sure how to get this solution with the presence of a fraction as a power in the question statement, looking for some help!

Answer 1

You can use $(x+3)^{-\frac{3}{5}}$ as a common term. Then you get:

$$(x+3)^{-\frac{3}{5}}(x+(x+3))=(x+3)^{-\frac{3}{5}}(2x+3)$$

It's because $\frac{(x+3)^{\frac{2}{5}}}{(x+3)^{-\frac{3}{5}}}=(x+3)^{\frac{3}{5}+\frac{2}{5}}=(x+3)$

Answer 2

$$x(x+3)^{-3/5} + (x+3)^{2/5}=x\color{blue}{(x+3)^{-3/5}}+(x+3)\color{blue}{(x+3)^{-3/5}}$$

$$=(x+x+3)\color{blue}{(x+3)^{-3/5}}=(2x+3)\color{blue}{(x+3)^{-3/5}}$$