How do you solve #2x^3+x^2-5x+2<=0# using a sign chart?

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Answer 1 · 2017-02-09T06:35:57

The answer is #x in ]-oo, 2] uu [1/2,1]#

Let #f(x)=2x^3+x^2-5x+2#

#f(1)=2+1-5+2=0#

Therefore, #(x-1)# is a factor of #f(x)#

To find the other factors, we do a long division

#color(white)(aaaa)##2x^3+x^2-5x+2##color(white)(aaaa)##|##x-1#

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

#color(white)(aaaaa)##0+3x^2-5x#

#color(white)(aaaaaaa)##+3x^2-3x#

#color(white)(aaaaaaaa)##+0-2x+2#

#color(white)(aaaaaaaaaaaa)##-2x+2#

#color(white)(aaaaaaaaaaaaa)##-0+0#

Therefore,

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

#=(x-1)(2x-1)(x+2)#

Now, we build the sign chart

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

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

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

#color(white)(aaaa)##x-1##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

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

Therefore,

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