To find the square of the given expression [tex]\((4 x \sqrt{5 x^2} + 2 x^2 \sqrt{6})^2\)[/tex], let's break it down and simplify step-by-step.
1. Understand the individual terms inside the expression:
[tex]\[
4 x \sqrt{5 x^2} + 2 x^2 \sqrt{6}
\][/tex]
2. Simplify [tex]\( \sqrt{5 x^2} \)[/tex]:
[tex]\[
\sqrt{5 x^2} = x \sqrt{5} \quad \text{(since \(x \geq 0\))}
\][/tex]
3. Substitute back into the expression:
[tex]\[
4 x \sqrt{5 x^2} = 4 x \cdot x \sqrt{5} = 4 x^2 \sqrt{5}
\][/tex]
So, the expression becomes:
[tex]\[
4 x^2 \sqrt{5} + 2 x^2 \sqrt{6}
\][/tex]
4. Factor out the common term [tex]\(2 x^2\)[/tex]:
[tex]\[
2 x^2 (2 \sqrt{5} + \sqrt{6})
\][/tex]
5. Square the entire expression:
[tex]\[
(2 x^2 (2 \sqrt{5} + \sqrt{6}))^2
\][/tex]
6. Apply the square to each part of the product:
[tex]\[
(2 x^2)^2 \cdot (2 \sqrt{5} + \sqrt{6})^2
\][/tex]
[tex]\[
4 x^4 \cdot (2 \sqrt{5} + \sqrt{6})^2
\][/tex]
7. Expand the binomial [tex]\((2 \sqrt{5} + \sqrt{6})^2\)[/tex]:
[tex]\[
(2 \sqrt{5} + \sqrt{6})^2 = (2 \sqrt{5})^2 + 2 \cdot (2 \sqrt{5}) \cdot (\sqrt{6}) + (\sqrt{6})^2
\][/tex]
[tex]\[
= 4 \cdot 5 + 2 \cdot 2 \sqrt{5} \cdot \sqrt{6} + 6
\][/tex]
[tex]\[
= 20 + 4 \sqrt{30} + 6
\][/tex]
[tex]\[
= 26 + 4 \sqrt{30}
\][/tex]
8. Combine these results:
[tex]\[
4 x^4 \cdot (26 + 4 \sqrt{30})
\][/tex]
9. Distribute [tex]\(4 x^4\)[/tex] through the terms inside the parentheses:
[tex]\[
4 x^4 \cdot 26 + 4 x^4 \cdot 4 \sqrt{30}
\][/tex]
[tex]\[
104 x^4 + 16 x^4 \sqrt{30}
\][/tex]
Therefore, the simplified form of [tex]\( (4 x \sqrt{5 x^2} + 2 x^2 \sqrt{6})^2 \)[/tex] is:
[tex]\[
104 x^4 + 16 x^4 \sqrt{30}
\][/tex]
Thus, the correct option is:
[tex]\[
104 x^4 + 16 x^4 \sqrt{30}
\][/tex]
So, the answer is the fourth option: [tex]\( \boxed{104 x^4 + 16 x^4 \sqrt{30}} \)[/tex].
