How do you solve # 3(4 - x)(2x + 1)>0#?

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Answer 1 · 2017-02-09T14:10:27

The answer is #x in ]-1/2, 4[#

Let #f(x)=3(4-x)(2x+1)#

We build a sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaa)##-1/2##color(white)(aaaa)##4##color(white)(aaaaa)##+oo#

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

#color(white)(aaaa)##4-x##color(white)(aaaaaaaa)##+##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 ]-1/2, 4[#