Question

How do you solve `x^3+7x^2-x-7<0`?

Answer 1

The answer is `x in ]-oo, -7 [uu]-1, 1[`

Explanation:

Let `f(x)=x^3+7x^2-x-7`

`f(1)=1+7-1-7=0`

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

To find the other factors, we do a long division

`color(white)(aaaa)` `x^3+7x^2-x-7` `color(white)(aaaa)` `|` `color(blue)(x-1)`

`color(white)(aaaa)` `x^3-x^2` `color(white)(aaaaaaaaaaaa)` `|` `color(red)(x^2+8x+7)`

`color(white)(aaaaa)` `0+8x^2-x`

`color(white)(aaaaaaa)` `+8x^2-8x`

`color(white)(aaaaaaaaa)` `+0+7x-7`

`color(white)(aaaaaaaaaaaaa)` `+7x-7`

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

Therefore,

`f(x)=(x-1)(x^2+8x+7)=(x-1)(x+1)(x+7)`

Now, we can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-7` `color(white)(aaaa)` `-1` `color(white)(aaaa)` `1` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `x+7` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

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

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

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

Therefore,

`f(x)<0` when `x in ]-oo, -7 [uu]-1, 1[`