Solve

$\frac{3x + 12}{x + 4} = (4x^2 + 8x + 18x^2)(2x)$

$3=(4x^2 + 8x + 18x^2)(2x)$

$3=(2x)(22x^2+8x)$

$3=44x^3+16x^2$

I'm afraid that's as far as I can get you.