1. **State the problem:** Solve the inequality $$\frac{(x - 1)(x - 2)}{(x + 1)(x - 4)} > 0$$. 2. **Recall the rule for rational inequalities:** The expression changes sign at zeros of numerator and denominator, i.e., at points where numerator or denominator is zero. 3. **Find critical points:** Numerator zeros: $x=1, 2$; Denominator zeros: $x=-1, 4$. 4. **Mark these points on the number line:** $-\infty < -1 < 1 < 2 < 4 < +\infty$. 5. **Test intervals between critical points:** - For $x < -1$, pick $x=-2$: numerator $( -2-1)(-2-2) = (-3)(-4) = 12 > 0$, denominator $(-2+1)(-2-4) = (-1)(-6) = 6 > 0$, fraction $>0$. - For $-1 < x < 1$, pick $x=0$: numerator $(0-1)(0-2) = (-1)(-2) = 2 > 0$, denominator $(0+1)(0-4) = (1)(-4) = -4 < 0$, fraction $<0$. - For $1 < x < 2$, pick $x=1.5$: numerator $(1.5-1)(1.5-2) = (0.5)(-0.5) = -0.25 < 0$, denominator $(1.5+1)(1.5-4) = (2.5)(-2.5) = -6.25 < 0$, fraction $>0$. - For $2 < x < 4$, pick $x=3$: numerator $(3-1)(3-2) = (2)(1) = 2 > 0$, denominator $(3+1)(3-4) = (4)(-1) = -4 < 0$, fraction $<0$. - For $x > 4$, pick $x=5$: numerator $(5-1)(5-2) = (4)(3) = 12 > 0$, denominator $(5+1)(5-4) = (6)(1) = 6 > 0$, fraction $>0$. 6. **Determine solution intervals where fraction > 0:** $$(-\infty, -1) \cup (1, 2) \cup (4, +\infty)$$ 7. **Exclude points where denominator is zero:** $x = -1, 4$ are excluded. **Final answer:** $$x \in (-\infty, -1) \cup (1, 2) \cup (4, +\infty)$$