1. **Problem:** Solve $a^3+a^2=36$ for $a$. 2. **Formula and idea:** First, move everything to one side: $$a^3+a^2-36=0$$ Then try factoring or testing simple integer values. 3. **Check for a factor:** Since the equation is a polynomial, we can test easy values. Try $a=3$: $$3^3+3^2=27+9=36$$ So $a=3$ works. 4. **Factor the expression:** Because $a=3$ is a root, $(a-3)$ is a factor. $$a^3+a^2-36=(a-3)(a^2+4a+12)$$ Now set each factor equal to zero: $$a-3=0$$ $$a^2+4a+12=0$$ 5. **Solve the factors:** From $a-3=0$, $$a=3$$ For $a^2+4a+12=0$, the discriminant is $$b^2-4ac=4^2-4(1)(12)=16-48=-32$$ Since this is negative, there are no real solutions from this factor. 6. **Final answer:** $$a=3$$