How do you solve #y^3-y^2-4y+4>0# using a sign chart?

1 Answer
Sep 20, 2016

Answer:

Solution is either #-2 < y < 1# or #y > 2#

Explanation:

Let us first factorize #y^3-y^2-4y+4#.

#y^3-y^2-4y+4=y^2(y-1)-4(y-1)=(y^2-4)(y-1)=(y+2)(y-2)(y-1)#

Hence we have to solve the inequality #y^3-y^2-4y+4>0# or #(y+2)(y-2)(y-1)>0#

From this we know that the product #(y+2)(y-2)(y-1)# is positive. It is apparent that sign of binomials #(y+2)#, #(y-2)# and #(y-1)# will change around the values #-2#. #2# and #1# respectively. In sign chart we divide the real number line using (y+2)(y-2)(y-1)these values, i.e. below #-2#, between #-2# and #1#, between #1# and #2# and above #2# and see how the sign of #(y+2)(y-2)(y-1)# changes.

Sign Chart

#color(white)(XXXXXXXXXXX)-2color(white)(XXXXX)1color(white)(XXXXX)2#

#(y+2)color(white)(XXXX)-ive color(white)(XXXX)+ive color(white)(XX)+ive color(white)(XXX)+ive#

#(y-2)color(white)(XXXX)-ive color(white)(XXXX)-ive color(white)(XX)-ive color(white)(XXX)+ive#

#(y-1)color(white)(XXXX)-ive color(white)(XXXX)-ive color(white)(XX)+ive color(white)(XXX)+ive#

#(y+2)(y-2)(y-1)#
#color(white)(XXXXXXXX)-ive color(white)(XXXX)+ive color(white)(XX)-ive color(white)(XXX)+ive#

It is observed that #(y+2)(y-2)(y-1) > 0# when either #-2 < y < 1# or #y > 2#, which is the solution for the inequality.