How do you solve $(x + 8) (x + 2) (x - 3) \geq 0$ $(x+8)(x+2)(x-3)>=0$ using a sign chart?

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How do you solve $(x + 8) (x + 2) (x - 3) \geq 0$ $(x+8)(x+2)(x-3)>=0$ using a sign chart?

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Answer 1 · 2016-12-19T19:01:24

The answer is $x \in [- 8, - 2] \cup [3, + \infty [$ $x in [-8,-2] uu [3, +oo[$

Explanation:

Let $f (x) = (x + 8) (x + 2) (x - 3)$ $f(x)=(x+8)(x+2)(x-3)$

Let's do the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 8$ $-8$ $a a a a$ $color(white)(aaaa)$ $- 2$ $-2$ $a a a a$ $color(white)(aaaa)$ $3$ $3$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 8$ $x+8$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x + 2$ $x+2$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x - 3$ $x-3$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

Therefore,

$f (x) \geq 0$ $f(x)>=0$ , when $x \in [- 8, - 2] \cup [3, + \infty [$ $x in [-8,-2] uu [3, +oo[$

graph{(x+8)(x+2)(x-3) [-154.2, 146, -60.4, 89.8]}

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How do you solve $(x + 8) (x + 2) (x - 3) \geq 0$ $(x+8)(x+2)(x-3)>=0$ using a sign chart?

1 Answer

Explanation:

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How do you solve $(x + 8) (x + 2) (x - 3) \geq 0$ $(x+8)(x+2)(x-3)>=0$ using a sign chart?