Question 339897
First factor,
{{{x^3+2x^2-3x>0}}}
{{{x(x^2+2x-3)>0}}}
{{{x(x+3)(x-1)>0}}}
Now break up the number line into 4 regions base on the zeros of the polynomial,
Region 1: ({{{-infinity}}},{{{-3}}})
Region 2: ({{{-3}}},{{{0}}})
Region 3: ({{{0}}},{{{1}}})
Region 4: ({{{1}}},{{{-infinity}}})
For each region choose a point in the region (not an endpoint).
Test the inequality.
If the inequality is satsfied, that region is part of the solution region.
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Region 1:{{{x=-4}}}
{{{x(x+3)(x-1)>0}}}
{{{-4(-4+3)(-4-1)>0}}}
{{{-4(-1)(-5)>0}}}
{{{-20>0}}}
False, this region is not part of the solution region.
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Region 2:{{{x=-2}}}
{{{x(x+3)(x-1)>0}}}
{{{-2(-2+3)(-2-1)>0}}}
{{{-2(1)(-3)>0}}}
{{{6>0}}}
True, this region is part of the solution region.
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Region 3:{{{x=1/2}}}
{{{x(x+3)(x-1)>0}}}
{{{(1/2)(1/2+3)(1/2-1)>0}}}
{{{1/2(7/2)(-1/2)>0}}}
{{{-7/8>0}}}
False, this region is not part of the solution region.
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Region 4:{{{x=2}}}
{{{x(x+3)(x-1)>0}}}
{{{2(2+3)(2-1)>0}}}
{{{2(5)(1)>0}}}
{{{10>0}}}
True, this region is part of the solution region.
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Solution Region :({{{-3}}},{{{0}}}) U ({{{1}}},{{{-infinity}}})
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Graphical verification: Look for the region where the function is above the x-axis.
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{{{drawing(300,300,-5,5,-5,5,blue(line(-3,10,-3,-10)),blue(line(0,10,0,-10)),blue(line(1,10,1,-10)),graph(300,300,-5,5,-5,5,x*(x+3)*(x-1)))}}}