Question

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

Answer 1

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

Explanation:

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}`