To find the product [tex]\( \left(4 x \sqrt{5 x^2}+2 x^2 \sqrt{6}\right)^2 \)[/tex], let's go through the detailed steps of expanding this expression.
Given:
[tex]\[
\left(4 x \sqrt{5 x^2}+2 x^2 \sqrt{6}\right)^2
\][/tex]
We'll denote [tex]\( A = 4 x \sqrt{5 x^2} \)[/tex] and [tex]\( B = 2 x^2 \sqrt{6} \)[/tex], so our expression becomes:
[tex]\[
(A + B)^2
\][/tex]
Using the algebraic identity for expanding a square of a sum, [tex]\((A + B)^2 = A^2 + 2AB + B^2\)[/tex], we get:
[tex]\[
(A + B)^2 = (4 x \sqrt{5 x^2})^2 + 2(4 x \sqrt{5 x^2})(2 x^2 \sqrt{6}) + (2 x^2 \sqrt{6})^2
\][/tex]
Now, we can square each term:
1. Squaring [tex]\( A \)[/tex]:
[tex]\[
(4 x \sqrt{5 x^2})^2 = (4 x)^2 \cdot (\sqrt{5 x^2})^2 = 16 x^2 \cdot 5 x^2 = 80 x^4
\][/tex]
2. Finding the product [tex]\( 2AB \)[/tex]:
[tex]\[
2 \cdot (4 x \sqrt{5 x^2}) \cdot (2 x^2 \sqrt{6}) = 2 \cdot 4 x \cdot 2 x^2 \cdot \sqrt{5 x^2} \cdot \sqrt{6}
= 8 x^3 \cdot \sqrt{30 x^2}
= 8 x^3 \cdot \sqrt{30 x^2}
= 8 x^3 \cdot \sqrt{30 x^2}
= 8 \sqrt{30} x^3 \cdot x
= 16 \sqrt{30} x^3 \cdot x
= 16 \sqrt{30 x^2} x^3 = 16 \sqrt{30} x^4
\][/tex]
3. Squaring [tex]\( B \)[/tex]:
[tex]\[
(2 x^2 \sqrt{6})^2 = 4 x^4 \cdot 6 = 24 x^4
\][/tex]
Adding all the parts together:
[tex]\[
80 x^4 + 16 \sqrt{30} x^4 + 24 x^4 = 104 x^4 + 16 \sqrt{30} x^4
\][/tex]
Thus, the expansion results in:
[tex]\[
\boxed{104 x^4 + 16 \sqrt{30} x^4}
\][/tex]
