Multiply the special product. $$ (\sqrt{7 r}+4 \sqrt{s})^{2}= $$ $$ \square $$

Question

UpStudy AI · Accepted Answer

To multiply the special product $$(\sqrt{7r} + 4\sqrt{s})^2$$, we can use the formula for the square of a binomial, which is given by:

$$
(a + b)^2 = a^2 + 2ab + b^2
$$

In this case, let $$a = \sqrt{7r}$$ and $$b = 4\sqrt{s}$$. Now we can calculate each term:

1. **Calculate $$a^2$$**:
   $$
   a^2 = (\sqrt{7r})^2 = 7r
   $$

2. **Calculate $$b^2$$**:
   $$
   b^2 = (4\sqrt{s})^2 = 16s
   $$

3. **Calculate $$2ab$$**:
   $$
   2ab = 2(\sqrt{7r})(4\sqrt{s}) = 8\sqrt{7rs}
   $$

Now, we can combine all these results together:

$$
(\sqrt{7r} + 4\sqrt{s})^2 = a^2 + 2ab + b^2 = 7r + 8\sqrt{7rs} + 16s
$$

Thus, the final result is:

$$
\boxed{7r + 8\sqrt{7rs} + 16s}
$$$$\boxed{7r + 8\sqrt{7rs} + 16s}$$

Multiply the special product. \[ (\sqrt{7 r}+4 \sqrt{s})^{2}= \] \( \square \)