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.
