$(\sqrt{63} + \sqrt{252}) \times (\sqrt{175} + \sqrt{28}) = $ $x$

Concept & Formula To multiply these binomials efficiently, first simplify all the surds (square roots) by factoring out the largest perfect square from each number. This will reveal like terms that can be added together before multiplying. $$ \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} $$ Step-by-Step Solution * **Given:** $(\sqrt{63} + \sqrt{252}) \times (\sqrt{175} + \sqrt{28}) = x$ * **Calculation:** Simplify the terms in the first bracket. Look for perfect square factors: $\sqrt{63} = \sqrt{9 \times 7} = 3\sqrt{7}$ $\sqrt{252} = \sqrt{36 \times 7} = 6\sqrt{7}$ First bracket sum: $3\sqrt{7} + 6\sqrt{7} = 9\sqrt{7}$ * Simplify the terms in the second bracket: $\sqrt{175} = \sqrt{25 \times 7} = 5\sqrt{7}$ $\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$ Second bracket sum: $5\sqrt{7} + 2\sqrt{7} = 7\sqrt{7}$ * Multiply the simplified brackets: $(9\sqrt{7}) \times (7\sqrt{7}) = (9 \times 7) \times (\sqrt{7} \times \sqrt{7})$ $= 63 \times 7$ $= 441$ Exam Strategy & Shortcut When you see a set of large numbers under square roots in a multiplication problem, assume they share a common irrational base. If you quickly spot that $\sqrt{28}$ is $2\sqrt{7}$ and $\sqrt{63}$ is $3\sqrt{7}$, you immediately know *all* terms are multiples of $\sqrt{7}$. Just divide the other numbers by 7 ($252 \div 7 = 36$, $175 \div 7 = 25$) to find their perfect square coefficients instantly. Common Pitfall A major trap is attempting to use the FOIL method (First, Outer, Inner, Last) on the unsimplified radicals. Multiplying $\sqrt{252} \times \sqrt{175}$ directly creates massive numbers that are incredibly difficult to simplify under exam time pressure. Always simplify surds first! Final Answer **Therefore, the correct answer is 441.**