To find the product [tex]\((2\sqrt{7} + 3\sqrt{6})(5\sqrt{2} + 4\sqrt{3})\)[/tex], we should use the distributive property of multiplication over addition (also known as the FOIL method in algebra). This gives us:
[tex]\[
(a + b)(c + d) = ac + ad + bc + bd
\][/tex]
Let's identify the terms:
- [tex]\(a = 2\sqrt{7}\)[/tex]
- [tex]\(b = 3\sqrt{6}\)[/tex]
- [tex]\(c = 5\sqrt{2}\)[/tex]
- [tex]\(d = 4\sqrt{3}\)[/tex]
Applying the distributive property step-by-step:
1. Calculate [tex]\(ac\)[/tex]:
[tex]\[
2\sqrt{7} \cdot 5\sqrt{2} = 2 \cdot 5 \cdot \sqrt{7 \cdot 2} = 10 \sqrt{14}
\][/tex]
2. Calculate [tex]\(ad\)[/tex]:
[tex]\[
2\sqrt{7} \cdot 4\sqrt{3} = 2 \cdot 4 \cdot \sqrt{7 \cdot 3} = 8 \sqrt{21}
\][/tex]
3. Calculate [tex]\(bc\)[/tex]:
[tex]\[
3\sqrt{6} \cdot 5\sqrt{2} = 3 \cdot 5 \cdot \sqrt{6 \cdot 2} = 15 \sqrt{12} = 15 \sqrt{4 \cdot 3} = 15 \cdot 2\sqrt{3} = 30 \sqrt{3}
\][/tex]
4. Calculate [tex]\(bd\)[/tex]:
[tex]\[
3\sqrt{6} \cdot 4\sqrt{3} = 3 \cdot 4 \cdot \sqrt{6 \cdot 3} = 12 \sqrt{18} = 12 \sqrt{9 \cdot 2} = 12 \cdot 3\sqrt{2} = 36 \sqrt{2}
\][/tex]
Now sum all the terms together:
[tex]\[
10\sqrt{14} + 8\sqrt{21} + 30\sqrt{3} + 36\sqrt{2}
\][/tex]
Thus, the product is:
[tex]\[
10\sqrt{14} + 8\sqrt{21} + 30\sqrt{3} + 36\sqrt{2}
\][/tex]
After evaluating and simplifying each term, we find that our solution matches:
[tex]\[
(10 \sqrt{14}, 8 \sqrt{21}, 30 \sqrt{3}, 36 \sqrt{2})
\][/tex]
Therefore, the correct answer from the given options is:
[tex]\[
10 \sqrt{14} + 8 \sqrt{21} + 30 \sqrt{3} + 36 \sqrt{2}
\][/tex]
So the answer is:
[tex]\[
\boxed{10 \sqrt{14} + 8 \sqrt{21} + 30 \sqrt{3} + 36 \sqrt{2}}
\][/tex]
