Expand (4 root 3 a + root 3 a^2)^2.

Step-by-step solution

1. 1

   Factor the common part out of the bracket before squaring. Both terms share $\sqrt3\,a$:

   $4\sqrt3\,a+\sqrt3\,a^{2}=\sqrt3\,a\left(4+a\right).$

   This is far cleaner than applying the $(u+v)^2$ formula to two surd terms.

2. 2

   Square the factored form. The square distributes over a product:

   $\left(\sqrt3\,a(a+4)\right)^{2}=\left(\sqrt3\right)^{2}a^{2}(a+4)^{2}=3a^{2}(a+4)^{2}.$

   Squaring is exactly what makes the radical disappear: $\left(\sqrt3\right)^2=3$.

3. 3

   Expand the remaining square.

   $(a+4)^{2}=a^{2}+8a+16.$

4. 4

   Multiply through by 3a^2.

   $3a^{2}\left(a^{2}+8a+16\right)=3a^{4}+24a^{3}+48a^{2}.$

5. 5

   Cross-check with the direct expansion. Using $(u+v)^2=u^2+2uv+v^2$ with $u=4\sqrt3 a$ and $v=\sqrt3 a^2$:

   $u^{2}=48a^{2},\qquad 2uv=2\cdot 4\sqrt3 a\cdot\sqrt3 a^{2}=24a^{3},\qquad v^{2}=3a^{4},$

   which sums to the same $3a^4+24a^3+48a^2$. Note the middle term also loses its radical, because $\sqrt3\cdot\sqrt3=3$.