F(x)= (4-x)(x+3)^2. Решите неравенство f'(x) >0.

$f(x)=(4-x)(x+3)^2;f'(x)=-(x+3)^2+2(4-x)(x+3)=$
$=- x^{2}-6x-9+8x+24-2 x^{2}-6x=-3 x^{2}-4x+15;$

0;-3 x^{2}-4x+15>0;" alt="f'(x)>0;-3 x^{2}-4x+15>0;" align="absmiddle" class="latex-formula">
$3 x^{2}+4x-15\ \textless \ 0;D_1=4+45=49;$
$x_1= \frac{-2-7}{3}=-3;x_2= \frac{-2+7}{3}= \frac{5}{3};$
__+__-3__-__ $\frac{5}{3}$ __+__
$(-3; \frac{5}{3})$