How do you solve $x^{3} + 3 x^{2} + x + 3 \leq 0$ $x^3+3x^2+x+3<=0$ ?

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Answer 1 · 2017-04-23T08:28:54

$x \leq 3$ $x<=3$

$x^{3} + 3 x^{2} + x + 3 \leq 0$ $x^3+3x^2+x+3<=0$

$\Leftrightarrow x^{2} (x + 3) + 1 (x + 3) \leq 0$ $hArrx^2(x+3)+1(x+3)<=0$

or $(x^{2} + 1) (x + 3) \leq 0$ $(x^2+1)(x+3)<=0$

As $x^{2} + 1$ $x^2+1$ is always positive,

dividing by $(x^{2} + 1)$ $(x^2+1)$ , we get $x + 3 \leq 0$ $x+3<=0$ or $x \leq - 3$ $x<=-3$

How do you solve x3+3x2+x+3≤0x^3+3x^2+x+3<=0?