1. **State the problem:** Solve the equation $$x^4 + 12x^2 + 35 = 0$$ for $x$. 2. **Rewrite the equation:** Notice this is a quadratic in terms of $x^2$. Let $y = x^2$, then the equation becomes: $$y^2 + 12y + 35 = 0$$ 3. **Factor the quadratic:** Find two numbers that multiply to 35 and add to 12. These are 7 and 5. $$y^2 + 7y + 5y + 35 = 0$$ Group terms: $$(y^2 + 7y) + (5y + 35) = 0$$ Factor each group: $$y(y + 7) + 5(y + 7) = 0$$ Factor out common binomial: $$(y + 7)(y + 5) = 0$$ 4. **Solve for $y$:** Set each factor equal to zero: $$y + 7 = 0 \Rightarrow y = -7$$ $$y + 5 = 0 \Rightarrow y = -5$$ 5. **Recall $y = x^2$:** So, $$x^2 = -7$$ $$x^2 = -5$$ 6. **Solve for $x$:** Since $x^2$ equals a negative number, solutions are complex: $$x = \pm \sqrt{-7} = \pm i\sqrt{7}$$ $$x = \pm \sqrt{-5} = \pm i\sqrt{5}$$ **Final answer:** $$x = \pm i\sqrt{7}, \quad x = \pm i\sqrt{5}$$