How do you solve and graph `x^3<=4x^2+3x`?

We solve this inequality with a sign chart.

`x^3<=4x^2+3x`

`x^3-4x^2-3x<=0`

`x(x^2-4x-3)<=0`

We need the roots of the quadratic equation

`x^2-4x-3=0`

The discrimenant is

`Delta=b^2-4ac=(-4)^2-4*(1)*(-3)=16+12=28`

As, `Delta>0`, there are 2 real roots

`x_2=(-b+sqrtDelta)/2=1/2(4+sqrt28)=2+sqrt7=4.646`

`x_1=(-b-sqrtDelta)/2=1/2(4+sqrt28)=2-sqrt7=-0.646`

Let `f(x)=x(x-x_1)(x-x_2)`

We can build the sign chart

`color(white)(aaaa)``x``color(white)(aaaa)``-oo``color(white)(aaaa)``x_1``color(white)(aaaa)``0``color(white)(aaaaaa)``x_2``color(white)(aaaa)``+oo`

`color(white)(aaaa)``x_1``color(white)(aaaaaaa)``-``color(white)(aaaa)``+``color(white)(aaaa)``+``color(white)(aaaa)``+`

`color(white)(aaaa)``x``color(white)(aaaaaaaa)``-``color(white)(aaaa)``-``color(white)(aaaa)``+``color(white)(aaaa)``+`

`color(white)(aaaa)``x_2``color(white)(aaaaaaa)``-``color(white)(aaaa)``-``color(white)(aaaa)``-``color(white)(aaaa)``+`

`color(white)(aaaa)``f(x)``color(white)(aaaaaa)``-``color(white)(aaaa)``+``color(white)(aaaa)``-``color(white)(aaaa)``+`

Therefore,

`f(x)<=0` when `x in (-oo, 2-sqrt7] uu [0, 2+sqrt7]`
graph{x^3-4x^2-3x [-12.34, 12.97, -9.01, 3.65]}