# Inequalities Interval Notation

posted by Sasha

Solve: (2 + x)(5 + x)(4 - x) < 0 (write answer in interval notation)

Which I think is xe (-infinity, -5) or (-2, 4)

and

Solve: (4x + 1)(3 - x) > or equal to 0 (write answer in interval notation)

Which im unsure

1. Jai

#1.
(2 + x)(5 + x)(4 - x) < 0
Equate each term inside the parenthesis to zero:
2 + x < 0
x < -2

5 + x < 0
x < -5

4 - x < 0
4 < x
x > 4

When x > 4, only the term (4 - x) is negative while the others are positive, so it satisfies the inequality.

When x < -5, say x = -6,
(2 - 6)(5 - 6)(4 + 6) < 0
-4 * -1 * 10 < 0
40 < 0
which is false. So x cannot be less than or equal to -5.

When x < -2 but greater than -5, say x = -3
(2 - 3)(5 - 3)(4 + 3) < 0
-1 * 2 * 7 < 0
-14 < 0
which is true. So -5 < x < -2. Combine it with x > 4, we'll have
(-5, -2) U (4, +infinity)

#2.
(4x + 1)(3 - x) ≥ 0
Same procedure we'll do. Equate each term to zero:
4x + 1 ≥ 0
4x ≥ -1
x ≥ -1/4

3 - x ≥ 0
3 ≥ x

We get the intersection. So x must be greater than or equal to -1/4 but less than or equal to 3:
[-1/4 , 3]

hope this helps~ `u`

2. Steve

After you get a bit of practice with this kind of stuff, you can feel confident using what you know about the graphs of polynomials.

(2 + x)(5 + x)(4 - x) < 0

You know this is a cubic, with three roots. The graph crosses the x-axis at each root. So, pick a very large negative value for x.

The product (-)(-)(+) is positive
So, you know that the graph crosses below the x-axis at -5 back up at -2 and back down at 4. So, f(x) < 0 on

(-5,-2)U(4,∞)

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