Answer
$x \lt 0~~$ or $~~1 \lt x \lt 3$
The solution set is $~~(-\infty, 0)\cup (1,3)$
Work Step by Step
$x^3+3x \lt 4x^2$
$x^3-4x^2+3x \lt 0$
$x(x^2-4x+3) \lt 0$
$(x)(x-3)(x-1) \lt 0$
When $~~x=0~~$, $~~x = 1,~~$ or $~~x=3,~~$ then $~~(x)(x-3)(x-1) = 0$
When $x \lt 0~~$ then $~~(x)(x-3)(x-1) \lt 0$
When $0 \lt x \lt 1,~~$ then $~~(x)(x-3)(x-1) \gt 0$
When $1 \lt x \lt 3,~~$ then $~~(x)(x-3)(x-1) \lt 0$
When $x \gt 3,~~$ then $~~(x)(x-3)(x-1) \gt 0$
Solution:
$x \lt 0~~$ or $~~1 \lt x \lt 3$
The solution set is $~~(-\infty, 0)\cup (1,3)$
