Question

In the given problem, solve and graph.

Express answers in both inequality and interval notaion.

$x^3\le 4x^2+3x$

Answer 1

$x^3\le 4x^2+3x$

$x^3-4x^2-3x\le 0$

$x(x^2-4x-3)\le 0$

$x=\frac{-(-4)\pm \sqrt{{(-4)}^2-4\cdot 1(-3)}}{2\cdot 1}:2\pm \sqrt{7}$

	x<$2-\sqrt{7}$	x=$2-\sqrt{7}$	$2-\sqrt{7}$<x<0	x=0	0<x<$2+\sqrt{7}$	x=$2+\sqrt{7}$	x>$2+\sqrt{7}$
x	-	-	-	0	+	+	+
$x^2-4x-3$	+	0	-	-	-	0	+
$x(x^2-4x-3)$	-	0	+	0	-	0	+

$x<2-\sqrt{7},0\le x\le 2+\sqrt{7}$

Interval Notation: $(-\infty ,2-\sqrt{7})\cup (0,2+\sqrt{7})$