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

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

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Answer 1 · 2016-10-24T13:42:02

The expression is $\geq 0$ $>=0$ for $x > 2$ $x>2$

Explanation:

let $f (x) = x^{3} + 2 x^{2} - 4 x - 8$ $f(x)=x^3+2x^2-4x-8$
Let us start by factorising the expression
By trial and error, · $(x - 2)$ $(x-2)$ is a factor as $f (2) = 0$ $f(2)=0$

let us make the long division
$x^{3} + 2 x^{2} - 4 x - 8$ $x^3+2x^2-4x-8$ $a a a$ $color(white)(aaa)$ $∣$ $∣$ $x - 2$ $x-2$
$x^{3} - 2 x^{2}$ $x^3-2x^2$ $a a a a a a a a a a a$ $color(white)(aaaaaaaaaaa)$ $∣$ $∣$ $x^{2} + 4 x + 4$ $x^2+4x+4$
$0 + 4 x^{2} - 4 x$ $0+4x^2-4x$
$a a a a$ $color(white)(aaaa)$ $4 x^{2} - 8 x$ $4x^2-8x$
$a a a a a$ $color(white)(aaaaa)$ $0 + 4 x - 8$ $0+4x-8$
$a a a a a a a a a$ $color(white)(aaaaaaaaa)$ $4 x - 8$ $4x-8$
$a a a a a a a a a a$ $color(white)(aaaaaaaaaa)$ $0 + 0$ $0+0$

so $f (x) = (x - 2) (x^{2} + 4 x + 4) = (x - 2) {(x + 2)}^{2}$ $f(x)=(x-2)(x^2+4x+4)=(x-2)(x+2)^2$
$f (x) \geq 0$ $f(x)>=0$
Looking at the factorisation
${(x + 2)}^{2} \geq 0$ $(x+2)^2>=0$ for all values of $x∈RR$
$(x-2)>=0$ for $x>=2$
So the answer is $x>=2$

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

1 Answer

Explanation:

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How do you solve x^3+2x^2-4x-8>=0x3+2x2−4x−8≥0x^3+2x^2-4x-8>=0 using a sign chart?

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