## Problem Solve for $a$ in the equation $a^3+a^2=36$. 1. Start with the given equation. $$a^3+a^2=36$$ 2. Move everything to one side (so it equals $0$). $$a^3+a^2-36=0$$ 3. Factor the polynomial by trying values that might multiply to $-36$. $$a^3+a^2-36=0$$ $$a^2(a+1)-36=0$$ 4. Try $a=3$: $$3^3+3^2=27+9=36$$ So $a=3$ is a solution, meaning $(a-3)$ is a factor. 5. Divide to factor completely. $$a^3+a^2-36=(a-3)(a^2+4a+12)$$ 6. Solve the quadratic $a^2+4a+12=0$ using the quadratic formula. $$a^2+4a+12=0$$ $$a=\frac{-4\pm\sqrt{4^2-4(1)(12)}}{2(1)}$$ $$a=\frac{-4\pm\sqrt{16-48}}{2}$$ $$a=\frac{-4\pm\sqrt{-32}}{2}$$ 7. Simplify the square root. $$\sqrt{-32}=\sqrt{32}\,i=\sqrt{16\cdot 2}\,i=4\sqrt{2}\,i$$ 8. Substitute back and simplify. $$a=\frac{-4\pm 4\sqrt{2}\,i}{2}$$ $$a=-2\pm 2\sqrt{2}\,i$$ ## Final answer $$a=3\quad\text{or}\quad a=-2\pm 2\sqrt{2}\,i$$