SOLUTION: The solve and graph the solution set (x +2)(x-20)(x+1)>0 {x|

SOLUTION: The solve and graph the solution set (x +2)(x-20)(x+1)>0 {x|_}

Question 470988: The solve and graph the solution set (x +2)(x-20)(x+1)>0 {x|_}
Found 2 solutions by jim_thompson5910, lwsshak3:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Solving

First, find the zeros of .

.

or or .

or or .

In ascending order, the zeros are: -2, -1, 20

So this means we have the following intervals to check: , , ,

Now in the interval , the expresssion is negative. Simply plug in any number less than -2 to see this (eg: plug in x=-3 to get (-3+2)*(-3-20)*(-3+1)=-46)

In the interval , the expresssion is positive. Simply plug in any number between -2 and -1 to see this (eg: plug in x=-1.5 to get (-1.5+2)*(-1.5-20)*(-1.5+1)=5.375)

In the interval , the expresssion is negative. Simply plug in any number between -1 and 20 to see this (eg: plug in x=10 to get (10+2)*(10-20)*(10+1)=-1320)

Finally, in the interval , the expresssion is positive. Simply plug in any number greater than 20 to see this (eg: plug in x=21 to get (21+2)*(21-20)*(21+1)=506)

So the following intervals yield positive outputs: and

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Answer:

The solution set is

Here's a graph to visually confirm this

Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
The solve and graph the solution set (x +2)(x-20)(x+1)>0
**
f(x)= (x +2)(x-20)(x+1)>0
Draw a number line with zeros on it:
-∞..........-2.......... -1...........20..........∞ (zeros on number line)
-∞.....-.....0......+.....0.....-......0....+.....∞ (sign of (f(x))
Solution is to find the sign of f(x) within intervals between zeros on the number line.
Here is the procedure:
Starting from the right side,
For x>20, f(x)>0
For -1 < x< 20, f(x)<0
For -2 < x< -1, f(x)>0
For -2 < x< -1, f(x)<0
For -∞ < x< -2, f(x)>0
You can notice that the sign switches every time we go thru a zero. This happens when the zero is an odd multiplicity (1, 3. 5, etc) Since zeros are of multiplicity 1, sign of f(x) switches every time we go thru a zero. When zeros are of even mulltiplicity (2, 4, 6, etc),The sign of f(x) does not change as we go thru the zero on the number line. Once you have the correct number line showing signs of f(x), you have the solution.
..
For given problem:
solution:(-2,-1) U (20,∞)
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