Multiply.
$7 \sqrt{5}(\sqrt{15}-4)$
Simplify your answer as much as possible.
The expression to simplify is \(7\sqrt{5}(\sqrt{15}-4)\). We recognize this as the product of a monomial \(7\sqrt{5}\) and a binomial \((\sqrt{15}-4)\). To simplify, we apply the distributive property, which states that \(a(b - c) = ab - ac\). We also need to recall the properties of square roots, specifically \(\sqrt{x} \cdot \sqrt{y} = \sqrt{x \cdot y}\) and the process for simplifying square roots by identifying perfect square factors.
We distribute the term \(7\sqrt{5}\) to each part of the binomial:
\[7\sqrt{5}(\sqrt{15}-4) = 7\sqrt{5} \cdot \sqrt{15} - 7\sqrt{5} \cdot 4.\]
Now, we simplify each term individually. For the first term, we multiply the numbers inside the radicals:
\[7\sqrt{5} \cdot \sqrt{15} = 7\sqrt{5 \cdot 15} = 7\sqrt{75}.\]
For the second term, we simply multiply the numerical coefficients:
\[7\sqrt{5} \cdot 4 = (7 \cdot 4)\sqrt{5} = 28\sqrt{5}.\]
Now, we need to simplify \(\sqrt{75}\) further. We look for perfect square factors of 75. Since \(75 = 25 \cdot 3\), we have:
\[\sqrt{75} = \sqrt{25 \cdot 3} = \sqrt{25} \cdot \sqrt{3} = 5\sqrt{3}.\]
Substituting this back into our expression for the first term:
\[7\sqrt{75} = 7 \cdot (5\sqrt{3}) = 35\sqrt{3}.\]
Finally, we combine the simplified terms to get the complete result:
\[35\sqrt{3} - 28\sqrt{5}.\]
The simplification of radicals is essential in many areas of physics and engineering, such as calculating the distance between two points in coordinate geometry when the coordinates involve square roots. For instance, if you are designing a path for a robot and need to calculate the exact distance between points \(A(3\sqrt{5}, 0)\) and \(B(0, \sqrt{15}-4)\) using the distance formula, you might encounter expressions similar to this one. Being able to simplify these expressions ensures that you can obtain an exact, clean result rather than dealing with complex decimals, which is crucial for precision.