To solve the product [tex]\((2 \sqrt{7} + 3 \sqrt{6})(5 \sqrt{2} + 4 \sqrt{3})\)[/tex], we need to use the distributive property, which is often referred to as the FOIL method for binomials. This states that:
[tex]\[
(a + b)(c + d) = ac + ad + bc + bd
\][/tex]
Applying this to our expression, we can break it down step-by-step:
1. Multiply [tex]\(2 \sqrt{7}\)[/tex] by [tex]\(5 \sqrt{2}\)[/tex]:
[tex]\[
2 \sqrt{7} \times 5 \sqrt{2} = 2 \times 5 \times \sqrt{7 \times 2} = 10 \sqrt{14}
\][/tex]
2. Multiply [tex]\(2 \sqrt{7}\)[/tex] by [tex]\(4 \sqrt{3}\)[/tex]:
[tex]\[
2 \sqrt{7} \times 4 \sqrt{3} = 2 \times 4 \times \sqrt{7 \times 3} = 8 \sqrt{21}
\][/tex]
3. Multiply [tex]\(3 \sqrt{6}\)[/tex] by [tex]\(5 \sqrt{2}\)[/tex]:
[tex]\[
3 \sqrt{6} \times 5 \sqrt{2} = 3 \times 5 \times \sqrt{6 \times 2} = 15 \sqrt{12}
\][/tex]
We need to simplify [tex]\(\sqrt{12}\)[/tex]:
[tex]\[
\sqrt{12} = \sqrt{4 \times 3} = 2 \sqrt{3}
\][/tex]
Thus,
[tex]\[
15 \sqrt{12} = 15 \times 2 \sqrt{3} = 30 \sqrt{3}
\][/tex]
4. Multiply [tex]\(3 \sqrt{6}\)[/tex] by [tex]\(4 \sqrt{3}\)[/tex]:
[tex]\[
3 \sqrt{6} \times 4 \sqrt{3} = 3 \times 4 \times \sqrt{6 \times 3} = 12 \sqrt{18}
\][/tex]
Simplify [tex]\(\sqrt{18}\)[/tex]:
[tex]\[
\sqrt{18} = \sqrt{9 \times 2} = 3 \sqrt{2}
\][/tex]
Thus,
[tex]\[
12 \sqrt{18} = 12 \times 3 \sqrt{2} = 36 \sqrt{2}
\][/tex]
Now, combine all the terms we have obtained:
[tex]\[
10 \sqrt{14} + 8 \sqrt{21} + 30 \sqrt{3} + 36 \sqrt{2}
\][/tex]
So, the expression simplifies to:
[tex]\[
10 \sqrt{14} + 8 \sqrt{21} + 30 \sqrt{3} + 36 \sqrt{2}
\][/tex]
Therefore, the correct answer is:
[tex]\[
\boxed{10 \sqrt{14} + 8 \sqrt{21} + 30 \sqrt{3} + 36 \sqrt{2}}
\][/tex]
