How do you solve $(x-3)^2(2x+1)<=0$ using a sign chart?

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Answer 1 · 2017-05-31T06:47:51

The solution is $x in (-oo,-1/2]uu {3}$

Let $f(x)=(x-3)^2(2x+1)$

The domain of $f(x)$ is $D_f(x)=RR$

$AA x in RR$ , $(x-3)^2>=0$

The sign chart is very simple

$color(white)(aaaa)$ $x$ $color(white)(aaaaaa)$ $-oo$ $color(white)(aaaa)$ $-1/2$ $color(white)(aaaa)$ $3$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $2x+1$ $color(white)(aaaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $(x-3)^2$ $color(white)(aaaaaa)$ $+$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

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

Therefore,

$f(x)<=0$ when $x in (-oo,-1/2]uu {3}$

How do you solve (x-3)^2(2x+1)<=0(x-3)^2(2x+1)<=0 using a sign chart?