Question 348753
x^3+x^2-20x<0
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Find the boundary values:
Solve x^3+x^2-20x = 0
x(x^2+x-20) = 0
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x(x+5)(x-4) = 0
x = -5 or x = 0 or x = 4
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To solve the inequality:
1st: Draw a number line and plot -5,0,4
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2nd: Test a check point from each interval
in the inequality to find the solution areas.
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x(x+5)(x-4) < 0

Test x = -10: you get -*-*- < 0 ; true
Test x = -2 : you get -*+*- < 0 ; false
Test x = 2  : you get +*+*- < 0 ; true
Test x = 10 ; you get +*+*+ < 0 ; false
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Solution: (-oo,-5)U(0,4)
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Cheers,
Stan H.
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