How do you solve #(x-1)(x+2)(2x-5)<0# using a sign chart?

1 Answer
Jan 17, 2017

Answer:

The answer is # x in ] -oo,-2 [ uu ] 1, 5/2[ #

Explanation:

Let #f(x)=(x-1)(x+2)(2x-5)#

Now, we can do our sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaa)##-2##color(white)(aaaa)##1##color(white)(aaaa)##5/2##color(white)(aaaaa)##+oo#

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

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

#color(white)(aaaa)##2x-5##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaa)##-##color(white)(aaaa)##+#

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

Therefore,

#f(x)<0# when # x in ] -oo,-2 [ uu ] 1, 5/2[ #