How do you solve $x^3+4x^2-x>=4$ using a sign chart?

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How do you solve $x^3+4x^2-x>=4$ using a sign chart?

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Answer 1 · 2016-11-17T12:16:25

The naswer is $x in [-4,-1] uu [1, +oo]$

Explanation:

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

Then, $f(1)=1+4-1-4=0$

Therefore, $(x-1)$ is a factor of $f(x)$

To find the other factors, let's do a long division

$color(white)(aaaaa)$ $x^3+4x^2-x-4$ $color(white)(aaaaa)$ $∣$ $x-1$

$color(white)(aaaaa)$ $x^3-x^2$ $color(white)(aaaaaaaaaaaaa)$ $∣$ $x^2+5x+4$

$color(white)(aaaaaa)$ $0+5x^2-x$

$color(white)(aaaaaaaa)$ $+5x^2-5x$

$color(white)(aaaaaaaaaaaa)$ $0+4x-4$

$color(white)(aaaaaaaaaaaaaa)$ $+4x-4$

$color(white)(aaaaaaaaaaaaaaa)$ $+0-0$

Therefore, $f(x)=(x-1)(x+1)(x+4)>=0$

Let's do the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-4$ $color(white)(aaaa)$ $-1$ $color(white)(aaaa)$ $+1$ $color(white)(aaaa)$ $+oo$

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

$color(white)(aaaa)$ $x+1$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-1$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

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

Therefore $f(x)>=0$ if $x in [-4,-1] uu [1, +oo]$

graph{x^3+4x^2-x-4 [-20.28, 20.27, -10.13, 10.13]}

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How do you solve $x^3+4x^2-x>=4$ using a sign chart?

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How do you solve $x^3+4x^2-x>=4$ using a sign chart?