1. **State the problem:** Solve the inequality $$x^4 + x^3 - x^2 - x > 0$$ and write the solution in interval form. 2. **Factor the expression:** Group terms to factor by grouping: $$x^4 + x^3 - x^2 - x = (x^4 + x^3) - (x^2 + x)$$ Factor out common terms in each group: $$= x^3(x + 1) - x(x + 1)$$ Factor out the common binomial factor $x + 1$: $$= (x + 1)(x^3 - x)$$ Further factor $x^3 - x$: $$x^3 - x = x(x^2 - 1) = x(x - 1)(x + 1)$$ So the full factorization is: $$x^4 + x^3 - x^2 - x = (x + 1)(x)(x - 1)(x + 1) = x(x - 1)(x + 1)^2$$ 3. **Rewrite the inequality:** $$x(x - 1)(x + 1)^2 > 0$$ 4. **Analyze the factors:** - The factor $(x + 1)^2$ is squared, so it is always $\geq 0$ and zero only at $x = -1$. - The sign of the product depends on $x$ and $x - 1$. 5. **Find critical points:** The critical points where the expression equals zero are: $$x = -1, 0, 1$$ 6. **Test intervals between critical points:** - Interval $(-2, -1)$: Choose $x = -2$ $$(-2)((-2) - 1)((-2) + 1)^2 = (-2)(-3)(-1)^2 = (-2)(-3)(1) = 6 > 0$$ - Interval $(-1, 0)$: Choose $x = -0.5$ $$(-0.5)((-0.5) - 1)((-0.5) + 1)^2 = (-0.5)(-1.5)(0.5)^2 = (-0.5)(-1.5)(0.25) = 0.1875 > 0$$ - Interval $(0, 1)$: Choose $x = 0.5$ $$0.5(0.5 - 1)(0.5 + 1)^2 = 0.5(-0.5)(1.5)^2 = 0.5(-0.5)(2.25) = -0.5625 < 0$$ - Interval $(1, 2)$: Choose $x = 2$ $$2(2 - 1)(2 + 1)^2 = 2(1)(3)^2 = 2(1)(9) = 18 > 0$$ 7. **Check points where expression equals zero:** - At $x = -1$, the factor $(x + 1)^2 = 0$, so expression equals zero. - At $x = 0$, expression equals zero. - At $x = 1$, expression equals zero. 8. **Write solution in interval form:** We want where expression is strictly greater than zero, so exclude points where it equals zero. From tests: $$(-\infty, -1) \cup (-1, 0) \cup (1, \infty)$$ **Final answer:** $$\boxed{(-\infty, -1) \cup (-1, 0) \cup (1, \infty)}$$