(x+5)(x-2)(x+1)>0
What is the solution set? Why am I not getting this at all?
easy way to do these ....
Since the expression is already factored, you are off to a good start.
the critical values are x = -5,-2 and 2
(those would have been the solution had it been an equation)
so we consider 4 sections:
1. x < -5
take any value in that domain, say x = -10
we get (-)(-)(-) < 0 ----- false
2. -5 < x < -1
take any value in that domain, say x = -4
(+)(-)(-) > 0 ----- true
3. -1 < x < 2
take any value of x in that domain, say x = 0
(+)(-)(+) < 0 ---- false
4. x > 2
let's pick x = 10
(+)(+)(+) > 0 ------ true
so solution:
-5 ≤ x ≤ -1 OR x > 2
third line contains a typo, should have said ...
the critical values are x = -5,-1 and 2
To find the solution set of the inequality (x+5)(x-2)(x+1) > 0, you need to understand the concept of polynomial inequalities. Here's a step-by-step explanation to help you understand:
Step 1: Factorize the polynomial.
Start by factoring the polynomial (x+5)(x-2)(x+1). This can be done by multiplying the terms together:
(x+5)(x-2)(x+1) = (x² + 4x + 5)(x+1)
= x³ + 5x² + x² + 4x + 5x + 5
= x³ + 6x² + 9x + 5
Step 2: Set the factors equal to zero.
Now, to determine the signs of the factors, we need to find where each factor becomes zero. We can do this by setting each factor equal to zero and solving for x:
x + 5 = 0 => x = -5
x - 2 = 0 => x = 2
x + 1 = 0 => x = -1
Step 3: Create a sign table.
To determine the sign of each factor within different intervals, create a sign table. This involves choosing test points within each interval and evaluating the sign of the polynomial at those points.
Interval (-∞, -5): Choose a test point, such as x = -6.
Substitute the test point into the polynomial:
(-6)³ + 6(-6)² + 9(-6) + 5 = -49 < 0
So, the sign of the polynomial in this interval is negative (-).
Interval (-5, -1): Choose a test point, such as x = -3.
Substitute the test point into the polynomial:
(-3)³ + 6(-3)² + 9(-3) + 5 = 17 > 0
So, the sign of the polynomial in this interval is positive (+).
Interval (-1, 2): Choose a test point, such as x = 0.
Substitute the test point into the polynomial:
0³ + 6(0)² + 9(0) + 5 = 5 > 0
So, the sign of the polynomial in this interval is positive (+).
Interval (2, ∞): Choose a test point, such as x = 3.
Substitute the test point into the polynomial:
3³ + 6(3)² + 9(3) + 5 = 101 > 0
So, the sign of the polynomial in this interval is positive (+).
Step 4: Determine the solution set.
The inequality (x+5)(x-2)(x+1) > 0 is asking for the values of x where the polynomial is greater than zero (positive). From the sign table, it can be observed that the polynomial is positive (+) in the intervals (-5, -1) and (2, ∞).
Thus, the solution set is (-5, -1) U (2, ∞).
It's important to understand that in polynomial inequalities, the signs of the factors determine the sign of the polynomial in different intervals. By analyzing the sign changes and evaluating the polynomial at test points, you can determine where the polynomial is positive or negative, leading to finding the solution set.