# How to re-write a solution to the inequalities as an "interval" that satisfies the inequality?I need to sketch the polynomial and determine the intervals that satisfy each inequality; a....

How to re-write a solution to the inequalities as an "interval" that satisfies the inequality?

I need to sketch the polynomial and determine the intervals that satisfy each inequality;

a. 2x³+3x²-17x-30 < 0

b. 3x⁴+x³-36x²+36x+16 ≧ 0

I've already gotten the roots and my teacher wrote that "All roots are fine" but I need to write the inequality as an interval. My solution to question a. was "Polynomials as f(x) = (x-3)(2x+3)(x+2) and Solution set x = {-2.5, 2, 3}

Any help is greatly appreciated! Thank you.

### 2 Answers | Add Yours

You need to sketch the graph of cubic polynomial `f(x) = 2x^3 + 3x^2 - 17x - 30` such that:

Notice that the curve intersects x axis at `x = -2.5, x = -2` and x`= 3` , hence, the polynomial has 3 real roots `x = -2.5, x = -2` and `x = 3` .

You need to consider only the negaive values of the cubic function, hence, you need to verify where the curve goes below x axis such that:

`x in (-oo,-2.5) => 2x^3 + 3x^2 - 17x - 30 < 0`

`x in (-2,3) => 2x^3 + 3x^2 - 17x - 30 < 0`

**Hence, the function takes negative values, `2x^3 + 3x^2 - 17x - 30 < 0` , if `x in (-oo,-2.5)U(-2,3).` **

I think that the ansmer to problem a would be x<-2 and -2<x<3. The equation in problem b factored is (x-2)(x-2)(3x+1)(x+4), and the intervals would be x=2, -4<=x<=-1/3.