Question

How do you solve and write the following in interval notation: `x^4+4x^3-21x^2>=0`?

Answer 1

`x le 7 uu x ge 3`

Explanation:

`x^4+4x^3-21x^2>=0->x^2(x^2+4x-21)`

but

`x^2 ge 0 forall x in RR` so
`x^2+4x-21=(x+7)(x-3) ge 0` or finally
`x le 7 uu x ge 3`

Answer 2

`(-oo,-7]uu[3,+oo)`

Explanation:

Factor this to find when the expression is equal to `0`. From there, we will create a sign chart to see when the expression is greater than `0`.

Factor `x^2` from each term.

`x^2(x^2+4x-21)>=0`

Factor with `7` and `-3`.

`x^2(x+7)(x-3)>=0`

Thus the expression equals `0` at `x=0`, `x=-7`, and `x=3`.

Use one test point in every interval between two pairs of zeros, and another test point between a zero and positive/negative infinities.

Before `x=-7`:

We can plug in `x=-8` into `x^2(x+7)(x-3)`.
`(-8)^2` is positive, `(-8+7)` is negative, and `(-8-3)` is negative, so their product will be positive. Thus the entire interval `(-oo,-7]` will be `>=0`.

Between `x=-7` and `x=0`:

Plugging in `x=-1` yields:

`overbrace((-1)^2)^+overbrace((-1+7))^(+)overbrace((-1-3))^(-) <=0`

This is negative, so the interval is not a part of the solution set.

Between `x=0` and `x=3`:

`overbrace((1)^2)^+overbrace((1+7))^(+)overbrace((1-3))^(-) <=0`

This is also negative, and not a part of the solution set.

From `x=3` to infinity:

Plugging in `x=4` yields:

`overbrace((4)^2)^+overbrace((4+7))^(+)overbrace((4-3))^+ >=0`

Since this is positive, `[3,+oo)` is also a part of the solution set.

Thus the entire solution is the union of the two positive intervals:

`(-oo,-7]uu[3,+oo)`