Question

How do you solve `x^3+7x^2+10x>=0`?

Answer 1

Solution is `-5 >= x>=-2` or `x >= 0`

Explanation:

Let us first factorize `x^3+7x^2+10x`.

`x^3+7x^2+10x=x(x^2+7x+10)=x(x^2+2x+5x+10)`

= `x(x(x+2)+5(x+2)=x(x+2)(x+5)`

Hence we have to solve the inequality `x^3+7x^2+10x>0` or `(x+5)(x+2)x>=0`

From this we know that for the product `(x+5)(x+2)x>=0`, signs of binomials `(x+5)`, `(x+2)` and `x` will change around the values `-5`. `-2` and `0` respectively. In sign chart we divide the real number line around these values, i.e. below `-5`, between `-5` and `-2`, between `-2` and `0` and above `0` and see how the sign of `(x+5)(x+2)x` changes.

Sign Chart

`color(white)(XXXXXXXXXXX)-5color(white)(XXXXX)-2color(white)(XXXXX)0`

`(x+5)color(white)(XXXX)-ive color(white)(XXXX)+ive color(white)(XX)+ive color(white)(XXX)+ive`

`(x+2)color(white)(XXXX)-ive color(white)(XXXX)-ive color(white)(XX)-ive color(white)(XXX)+ive`

`xcolor(white)(XXXXXXX)-ive color(white)(XXXX)-ive color(white)(XX)+ive color(white)(XXX)+ive`

`(x+5)(x+2)x`
`color(white)(XXXXXXXX)-ive color(white)(XXXX)+ive color(white)(XX)-ive color(white)(XXX)+ive`

It is observed that `(x+5)(x+2)x>= 0` when either `-5 >= x>=-2` or `x >= 0`, which is the solution for the inequality.