Question

How do you solve `2x^3+5x^2>6x+9` using a sign chart?

Answer 1

The answer is `x in ] -3, -1 [ uu ]3/2, +oo[`

Explanation:

Let's rewrite the inequality as

`2x^3+5x^2-6x-9>0`

We must factorise the LHS

Let `f(x)=2x^3+5x^2-6x-9`

We calculate

`f(-1)=-2+5+6-9=0`

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

To find the other factors, we do a long division

`color(white)(aaaa)` `2x^3+5x^2-6x-9` `color(white)(aaaa)` `∣` `x+1`

`color(white)(aaaa)` `2x^3+2x^2` `color(white)(aaaaaaaaaaaa)` `∣` `2x^2+3x-9`

`color(white)(aaaaaa)` `0+3x^2-6x`

`color(white)(aaaaaaaa)` `+3x^2+3x`

`color(white)(aaaaaaaaaa)` `+0-9x-9`

`color(white)(aaaaaaaaaaaaaa)` `-9x-9`

`color(white)(aaaaaaaaaaaaaaa)` `-0-0`

Therefore,

`(2x^3+5x^2-6x-9)/(x+1)=2x^2+3x-9=(2x-3)(x+3)`

So,

`2x^3+5x^2-6x-9=(x+1)(2x-3)(x+3)>0`

Now we can do the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaa)` `-3` `color(white)(aaaa)` `-1` `color(white)(aaaa)` `3/2` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+3` `color(white)(aaaaaa)` `-` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

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

`color(white)(aaaa)` `2x-3` `color(white)(aaaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

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

Therefore,

`f(x)>0` when `x in ] -3, -1 [ uu ]3/2, +oo[`