x^3(2x-7)^2(x+5)=0
I don't know how to solve this and I need this part so I can find the local max and min.
$\displaystyle
x^3(x+5)(2x-7)^2=0
$
$\displaystyle f(x)=(x^4+5x^3)(2x-7)^2=0$
differentiate wrt x using product rule,
$\displaystyle (x^4+5x^3)2(2x-7)+(4x^3+15x^2)(2x-7)^2=0$
$\displaystyle x^2(2x-7)[2(x^2+5x)+(4x+15)(2x-7)]$
$\displaystyle x^2(2x-7)(2x^2+10x+8x^2+2x-105)$
$\displaystyle f'(x)=x^2(2x-7)(10x^2+12x-105)=0$