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Solve each inequality algebraically.

x2(3+x)(x+4)(x+5)(x-1)≥0

Short Answer

Expert verified

Solution to the inequalityx2(3+x)(x+4)(x+5)(x-1)≥0 is (-∞,-5)∪[-4,-3]∪{0}∪(1,∞).

Step by step solution

01

Step 1. Given information  

We have been given an inequality x2(3+x)(x+4)(x+5)(x-1)≥0.

We have to solve this inequality algebraically.

02

Step 2. Determine the real numbers at which the expression f equals zero and at which the expression f is undefined.  

Assume f(x)=x2(3+x)(x+4)(x+5)(x-1).

x2=0x=03+x=0x=-3x+4=0x=-4x+5=0x=-5x-1=0x=1

03

Step 3. Form the intervals  

Using the values of x found in previous step, we can divide the real numbers in the intervals:

(-∞,-5)∪(-5,-4)∪(-4,-3)∪(-3,0)∪(0,1)∪(1,∞)

04

Step 4. Select a number in each interval and evaluate f at the number  

Create the following table:

Interval(-∞,-5)
(-5,-4)
(-4,-3)
(-3,0)
(0,1)
(1,∞)
Number chosen-6
-4.5
-3.5
-1
0.5
2
Value of ff(-6)=2167
f(-4.5)=-5.5
f(-3.5)=0.45
f(-1)=-0.75
f(0.5)=-1.43
f(2)=1207
Conclusionpositivenegative
positive
negative
negativepositive
05

Step 5. Identify the interval 

Since we want to know where f is positive or zero, we conclude that f(x)≥0in the interval (-∞,-5)∪[-4,-3]∪{0}∪(1,∞).

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