How do you solve $x^3+2x^2<=8x$ using a sign chart?

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How do you solve $x^3+2x^2<=8x$ using a sign chart?

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Answer 1 · 2017-02-21T10:55:12

The solution is $x in ]-oo,-4]uu[0, 2]$

Explanation:

Let's rewrite and factorise the inequality

$x^3+2x^2<=8x$

$x^3+2x^2-8x<=0$

$x(x^2+2x-8)<=0$

$x(x-2)(x+4)<=0$

Let $f(x)=x(x-2)(x+4)$

Now, we can build the sign chart

$color(white)(aaaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-4$ $color(white)(aaaa)$ $0$ $color(white)(aaaaa)$ $2$ $color(white)(aaaaa)$ $+oo$

$color(white)(aaaa)$ $x+4$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x$ $color(white)(aaaaaaaaa)$ $-$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-2$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

Therefore,

$f(x)<=0$ , when $x in ]-oo,-4]uu[0, 2]$

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1 Answer

Explanation:

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