How do you solve $x^{3} + 2 x^{2} - 4 x - 8 \geq 0$ $x^3+2x^2-4x-8>=0$ ?

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How do you solve $x^{3} + 2 x^{2} - 4 x - 8 \geq 0$ $x^3+2x^2-4x-8>=0$ ?

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Answer 1 · 2016-11-23T09:12:26

The answer is $x \geq 2$ $x>=2$ or $x \in [2, + \infty [$ $x in [2,+oo[$

Explanation:

Let $f (x) = x^{3} + 2 x^{2} - 4 x - 8$ $f(x)=x^3+2x^2-4x-8$

Then $f (2) = 8 + 8 - 8 - 8 = 0$ $f(2)=8+8-8-8=0$
Therefore, $(x - 2)$ $(x-2)$ is a factor of $f (x)$ $f(x)$

$f (- 2) = - 8 + 8 + 8 - 8 = 0$ $f(-2)=-8+8+8-8=0$
Therefore, $(x + 2)$ $(x+2)$ is a factor of $f (x)$ $f(x)$

So, $(x + 2) (x - 2) = x^{2} - 4$ $(x+2)(x-2)=x^2-4$ is a factor of $f (x)$ $f(x)$

To find the last factor, let's do a long division

$a a a a$ $color(white)(aaaa)$ $x^{3} + 2 x^{2} - 4 x - 8$ $x^3+2x^2-4x-8$ $a a a a$ $color(white)(aaaa)$ $∣$ $∣$ $x^{2} - 4$ $x^2-4$

$a a a a$ $color(white)(aaaa)$ $x^{3}$ $x^3$ $a a a a a a a$ $color(white)(aaaaaaa)$ $- 4 x$ $-4x$ $a a a a a a a$ $color(white)(aaaaaaa)$ $∣$ $∣$ $x + 2$ $x+2$

$a a a a$ $color(white)(aaaa)$ $0 + 2 x^{2}$ $0+2x^2$ $a a a a a$ $color(white)(aaaaa)$ $0 - 8$ $0-8$

$a a a a a a$ $color(white)(aaaaaa)$ $+ 2 x^{2}$ $+2x^2$ $a a a a a a a$ $color(white)(aaaaaaa)$ $- 8$ $-8$

$a a a a a a a a$ $color(white)(aaaaaaaa)$ $0$ $0$ $a a a a a a a a a a$ $color(white)(aaaaaaaaaa)$ $0$ $0$

So, $f (x) = {(x + 2)}^{2} (x - 2)$ $f(x)=(x+2)^2(x-2)$

As ${(x + 2)}^{2} > 0$ $(x+2)^2>0$

Therefore, the sign of $f (x)$ $f(x)$ will depend on the sign of $(x - 2)$ $(x-2)$

When $x < 2$ $x<2$ , $f (x) < 0$ $f(x)<0$

and when $x \geq 2$ $x>=2$ , $f (x) \geq 0$ $f(x)>=0$

graph{x^3+2x^2-4x-8 [-20.28, 20.27, -10.14, 10.14]}

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How do you solve $x^{3} + 2 x^{2} - 4 x - 8 \geq 0$ $x^3+2x^2-4x-8>=0$ ?

1 Answer

Explanation:

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