1. State the problem: Solve for $a$ in $a^3+a^2=36$. 2. Start with the given equation: $$a^3+a^2=36$$ 3. Factor the left side by taking out the common factor $a^2$: $$a^3+a^2=a^2(a+1)$$ 4. Rewrite the equation using the factorization: $$a^2(a+1)=36$$ 5. Bring everything to one side to look for roots: $$a^2(a+1)-36=0$$ $$a^3+a^2-36=0$$ 6. Try factoring the cubic. Test $a=3$: $$3^3+3^2-36=27+9-36=0$$ So $(a-3)$ is a factor. 7. Divide the cubic by $(a-3)$ (synthetic division): Coefficients: $1,\ 1,\ 0,\ -36$. Bring down $1$. Multiply by $3$ to get $3$, add to $1$ to get $4$. Multiply by $3$ to get $12$, add to $0$ to get $12$. Multiply by $3$ to get $36$, add to $-36$ to get $0$. So the quotient is $a^2+4a+12$. 8. Write the factored form: $$a^3+a^2-36=(a-3)(a^2+4a+12)$$ 9. Set each factor equal to $0$ and solve: $$a-3=0 \Rightarrow a=3$$ 10. For the quadratic, use the discriminant $\Delta$: $$a^2+4a+12=0$$ $$\Delta=4^2-4\cdot1\cdot12=16-48=-32$$ 11. Because $\Delta<0$, the quadratic has no real solutions (it gives complex solutions only). 12. Final answer (real solutions): $a=3$.