How do you solve #(x-3)^2(2x+1)<=0# using a sign chart?

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Answer 1 · 2017-05-31T06:47:51

The solution is #x in (-oo,-1/2]uu {3}#

Let #f(x)=(x-3)^2(2x+1)#

The domain of #f(x)# is #D_f(x)=RR#

#AA x in RR#, #(x-3)^2>=0#

The sign chart is very simple

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaa)##-1/2##color(white)(aaaa)##3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##2x+1##color(white)(aaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##(x-3)^2##color(white)(aaaaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

Therefore,

#f(x)<=0# when #x in (-oo,-1/2]uu {3}#