Question

How do you solve `2x^3+x^2-5x+2<=0` using a sign chart?

Answer 1

The answer is `x in ]-oo, 2] uu [1/2,1]`

Explanation:

Let `f(x)=2x^3+x^2-5x+2`

`f(1)=2+1-5+2=0`

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

To find the other factors, we do a long division

`color(white)(aaaa)` `2x^3+x^2-5x+2` `color(white)(aaaa)` `|` `x-1`

`color(white)(aaaa)` `2x^3-2x^2` `color(white)(aaaaaaaaaaa)` `|` `2x^2+3x-2`

`color(white)(aaaaa)` `0+3x^2-5x`

`color(white)(aaaaaaa)` `+3x^2-3x`

`color(white)(aaaaaaaa)` `+0-2x+2`

`color(white)(aaaaaaaaaaaa)` `-2x+2`

`color(white)(aaaaaaaaaaaaa)` `-0+0`

Therefore,

`f(x)=(x-1)(2x^2+3x-2)`

`=(x-1)(2x-1)(x+2)`

Now, we build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaa)` `-oo` `color(white)(aaaa)` `-2` `color(white)(aaaa)` `1/2` `color(white)(aaaa)` `1` `color(white)(aaaaaa)` `+oo`

`color(white)(aaaa)` `x+2` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `2x-1` `color(white)(aaaaa)` `-` `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)` `f(x)` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(x)<=0` when `x in ]-oo, 2] uu [1/2,1]`