Answer
Solutions:
$x=0$
$x=-\frac{2}{3}$
$x=+\frac{2}{3}$
$x=-3$
Work Step by Step
$9x^4+27x^3-4x^2-12x=0$
$9x^3(x+3)-4x(x+3)=0$
$(9x^3-4x)(x+3)=0$
$x(9x^2-4)(x+3)=0$
$x[(3x)^2-2^2](x+3)=0$
$x(3x+2)(3x-2)(x+3)=0$
Solutions:
$x=0$
$3x+2=0$
$x=-\frac{2}{3}$
$3x-2=0$
$x=+\frac{2}{3}$
$x+3=0$
$x=-3$
Check the solutions:
$x=0$:
$9(0)^4+27(0)^3-4(0)^2-12(0)=0$
$x=-\frac{2}{3}$
$9(-\frac{2}{3})^4+27(-\frac{2}{3})^3-4(-\frac{2}{3})^2-12(-\frac{2}{3})=9\frac{16}{81}-27\frac{8}{27}-4\frac{4}{9}+8=0$
$x=\frac{2}{3}$
$9(\frac{2}{3})^4+27(\frac{2}{3})^3-4(\frac{2}{3})^2-12(\frac{2}{3})=9\frac{16}{81}+27\frac{8}{27}-4\frac{4}{9}-8=0$
$x=-3$:
$9(-3)^4+27(-3)^3-4(-3)^2-12(-3)=9(81)-27(27)-4(9)+36=0$
