Question 338692
{{{x^2+8x+7>0}}}
{{{(x+1)(x+7)>0}}}
Break up the number line into 3 regions.
Region 1: ({{{-infinity}}},{{{-7}}})
Region 2: ({{{-7}}},{{{-1}}})
Region 3: ({{{-1}}},{{{infinity}}})
Choose a point in each region (not an endpoint).
Test the inequality.
If the inequality is satisfied, the region is part of the solution region.
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Region 1: {{{x=-8}}}
{{{(x+1)(x+7)>0}}}
{{{(-8+1)(-8+7)>0}}}
{{{(-7)(-1)>0}}}
{{{7>0}}}
True, this region is part of the solution.
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Region 2: {{{x=-2}}}
{{{(x+1)(x+7)>0}}}
{{{(-2+1)(-2+7)>0}}}
{{{(-1)(5)>0}}}
{{{-5>0}}}
False, this region is not part of the solution.
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Region 3: {{{x=0}}}
{{{(x+1)(x+7)>0}}}
{{{(1)(7)>0}}}
{{{7>0}}}
True, this region is part of the solution.
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Solution: ({{{-infinity}}},{{{-7}}}) U ({{{-1}}},{{{infinity}}})
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{{{x^3-x^2-6x>0}}}
{{{x(x^2-x-6)>0}}}
{{{x(x-3)(x+2)>0}}}
Break up the number line into 4 regions.
Region 1: ({{{-infinity}}},{{{-2}}})
Region 2: ({{{-2}}},{{{0}}})
Region 3: ({{{0}}},{{{3}}})
Region 4: ({{{3}}},{{{infinity}}})
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Follow the same procedure as outlined above to find the solution region.