How do you solve and write the following in interval notation: #(x^2-5x+6)/(x+3) >0#?

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Answer 1 · 2017-10-11T12:24:04

The solution is #x in (-3,2) uu (3,+oo)#

We cannot do crossing over

Factorise the inequality

#(x^2-5x+6)/(x+3)>0#

#((x-2)(x-3))/(x+3)>0#

Let #f(x)=((x-2)(x-3))/(x+3)#

Build the sign chart

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

#color(white)(aaaa)##x+3##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

#color(white)(aaaa)##x-3##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

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

Therefore,

#f(x)>0# when #x in (-3,2) uu (3,+oo)#

graph{(x^2-5x+6)/(x+3) [-148.4, 152, -74.4, 76]}

graph{(x^2-5x+6)/(x+3) [-0.486, 4.993, -1.132, 1.61]}