To find the product [tex]\((2 \sqrt{7} + 3 \sqrt{6})(5 \sqrt{2} + 4 \sqrt{3})\)[/tex], we will apply the distributive property, also known as the FOIL method, which stands for First, Outer, Inner, and Last terms in binomial multiplication.
First, let's distribute each term in the first binomial to each term in the second binomial:
[tex]\[
(2 \sqrt{7} + 3 \sqrt{6})(5 \sqrt{2} + 4 \sqrt{3}) = (2 \sqrt{7} \cdot 5 \sqrt{2}) + (2 \sqrt{7} \cdot 4 \sqrt{3}) + (3 \sqrt{6} \cdot 5 \sqrt{2}) + (3 \sqrt{6} \cdot 4 \sqrt{3})
\][/tex]
Now, evaluate each term step by step:
1. First term:
[tex]\[
2 \sqrt{7} \cdot 5 \sqrt{2} = (2 \cdot 5) \cdot (\sqrt{7} \cdot \sqrt{2}) = 10 \sqrt{14}
\][/tex]
2. Outer term:
[tex]\[
2 \sqrt{7} \cdot 4 \sqrt{3} = (2 \cdot 4) \cdot (\sqrt{7} \cdot \sqrt{3}) = 8 \sqrt{21}
\][/tex]
3. Inner term:
[tex]\[
3 \sqrt{6} \cdot 5 \sqrt{2} = (3 \cdot 5) \cdot (\sqrt{6} \cdot \sqrt{2}) = 15 \sqrt{12} = 15 \sqrt{4 \cdot 3} =15 \cdot 2 \sqrt{3} = 30 \sqrt{3}
\][/tex]
4. Last term:
[tex]\[
3 \sqrt{6} \cdot 4 \sqrt{3} = (3 \cdot 4) \cdot (\sqrt{6} \cdot \sqrt{3}) = 12 \sqrt{18} = 12 \sqrt{9 \cdot 2} =12 \cdot 3 \sqrt{2}= 36 \sqrt{2}
\][/tex]
Collect all the terms together:
[tex]\[
10 \sqrt{14} + 8 \sqrt{21} + 30 \sqrt{3} + 36 \sqrt{2}
\][/tex]
So the correct answer is:
[tex]\[
10 \sqrt{14} + 8 \sqrt{21} + 30 \sqrt{3} + 36 \sqrt{2}
\][/tex]
Which matches option:
[tex]\[
\boxed{10 \sqrt{14} + 8 \sqrt{21} + 30 \sqrt{3} + 36 \sqrt{2}}
\][/tex]
