Question 1090338
No, break up the number line into distinct regions using the critical values of the inequality (values that make the left hand side equal zero).
So,
{{{x=2}}}
{{{x=-1}}}
{{{x=5}}}
So then the four regions are:
Region 1: {{{x<-1}}}
Region 2: {{{-1<x<2}}}
Region 3: {{{2<x<5}}}
Region 4: {{{x>5}}}
For each region, choose a value in the region (not an endpoint), test the inequality, determine if that point is part of the solution.
Region 1: {{{x<-1}}}
{{{x=-2}}}
{{{(x-2)^2*(x+1)^3*(x-5)<=0}}}
{{{(-2-2)^2*(-2+1)^3*(-2-5)<=0}}}
{{{(-4)^2*(-1)^3*(-7)<=0}}}
{{{(16)*(-1)*(-7)<=0}}}
{{{112<=0}}}
False, not part of the solution.
Region 2: {{{-1<x<2}}}
{{{x=0}}}
{{{(x-2)^2*(x+1)^3*(x-5)<=0}}}
{{{(0-2)^2*(0+1)^3*(0-5)<=0}}}
{{{(4)*(1)*(-5)<=0}}}
{{{-5<=0}}}
True, part of the solution.
Region 3: {{{2<x<5}}}
{{{x=3}}}
{{{(x-2)^2*(x+1)^3*(x-5)<=0}}}
{{{(3-2)^2*(3+1)^3*(3-5)<=0}}}
{{{(1)*(64)*(-2)<=0}}}
{{{-128<=0}}}
True, part of the solution.
Region 4: {{{x>5}}}
{{{x=6}}}
{{{(x-2)^2*(x+1)^3*(x-5)<=0}}}
{{{(6-2)^2*(6+1)^3*(6-5)<=0}}}
{{{(16)*(343)*(1)<=0}}}
{{{5488<=0}}}
False, not part of the solution.
So then, the solution region is
{{{-1<x<2}}}U{{{2<x<5}}}
The inequality also holds when {{{x=2}}} because of the equal sign in the inequality so,
{{{-1<=x<=5}}}
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*[illustration fer19.JPG].