(1) [tex]x^3-x^2-4x+4<0[/tex]
with this type of polynomials I first try to find one root by guess: it is 1. Then divide the polynomial by that root, (x-1), which quickly reveals the remaining roots:
[tex](x^3-x^2-4x+4)/(x-1)=x^2-4=(x-2)(x+2)[/tex]
Se the cubic polynomial intersects the x axis in three points {-2,1,2}
Because the coefficient of the highest power (3) is positive and the power is odd, the expression is negative for x<-2, then it is positive until it crosses x again at x=1, then negative between 1 and 2, and finally becomes positive for x>2.
So the solutions are:
[tex]x^3-x^2-4x+4<0\\(x+2)(x-1)(x-2)<0\implies -\infty < x<-2\,\,\,\mbox{and}\,\,\,1<x<2[/tex]
(2) In this case start with a substitution: [tex]z\leftarrow x^2[/tex]
which will transform the quartic expression to a quadratic one:
[tex]x^4-25x^2+144\rightarrow z^2-25z+144\implies z_1=16, z_2=9\\\implies x_{1,2}=\pm 4, \,\,x_{3,4}=\pm3\\(x+4)(x+3)(x-3)(x-4)<0[/tex]
By a similar argument as with (1) you can determine the positive and negative regions. Since the highest power is even, the region x<-4 is in the positive, then crosses into negative for -4<x<-3, positive for -3<x<3, negative for 3<x<4, and finally positive for 4<x. So the solutions of the inequality are:
{-4<x<-3, 3<x<4}
