How do you solve $x^3+x^2+4x+4>0$ using a sign chart?

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Answer 1 · 2017-02-04T11:23:48

The answer is $x in ]-1, +oo[$

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

Before, we need to find the factors of $f(x)$

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

Therefore, $(x+1)$ is a factor

To find the other factors, we perform a long division

$color(white)(aaaa)$ $x^3+x^2+4x+4$ $color(white)(aaaa)$ $|$ $x+1$

$color(white)(aaaa)$ $x^3+x^2$ $color(white)(aaaaaaaaaaaa)$ $|$ $x^2+4$

$color(white)(aaaaa)$ $0+0+4x+4$

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

$color(white)(aaaaaaaaaaaaa)$ $0+0$

Therefore,

$f(x)=(x+1)(x^2+4)$

$AA x in RR,(x^2+4)>0$

So, we can build the sign chart

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

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

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

Therefore,

$f(x)>0$ when $x in ]-1, +oo[$

How do you solve x^3+x^2+4x+4>0x^3+x^2+4x+4>0 using a sign chart?