Rewrite $ \frac{2}{5}(x-2)^\frac{5}{2} + \frac{4}{3}(x-2)^\frac{3}{2} + c$ to $\frac{2}{15}(3x+4)(x-2)^\frac{3}{2} + c $

Question

I must get the expression $ \frac{2}{5}(x-2)^\frac{5}{2} + \frac{4}{3}(x-2)^\frac{3}{2} + c$

into $$ \frac{2}{15}(3x+4)(x-2)^\frac{3}{2} + c $$

I tried expanding the $\frac{4}{3}(x-2)^\frac{3}{2}$ into $\frac{4}{3}(x-2)^1(x-2)^\frac{1}{2}$, and then distributed the $\frac{4}{3}$ to get $(\frac{4}{3}x - \frac{8}{3})(x-2)^\frac{1}{2}$, but I do not see how to get on from there.

Answer 1

$ \frac{2}{5}(x-2)^\frac{5}{2} + \frac{4}{3}(x-2)^\frac{3}{2} + c$

$ \frac{2}{5}(x-2)^\frac{3}{2}(x-2) + \frac{4}{3}(x-2)^\frac{3}{2} + c$

$(x-2)^\frac{3}{2}( \frac{2}{5}(x-2) + \frac{4}{3}) + c$

$(x-2)^\frac{3}{2}(\frac{6}{15}x-\frac{8}{15}) + c$

$(x-2)^\frac{3}{2}(\frac{2\times3x}{15}-\frac{2\times4}{15}) + c$

$ \frac{2}{15}(3x+4)(x-2)^\frac{3}{2} + c $