Area tiles show \(t^2+7t+12\). Which rectangle side lengths are correct?

Correct answer: \((t+3)\) and \((t+4)\)

The correct answer is B: \((t+3)\) and \((t+4)\). To arrange the area tiles into a rectangle, find two numbers whose product is the constant term 12 and whose sum is the coefficient of \(t\), namely 7. The numbers 3 and 4 satisfy both conditions: \(3\times4=12\) and \(3+4=7\). Therefore, \(t^2+7t+12=(t+3)(t+4)\). On multiplying, the four regions are \(t^2\), \(4t\), \(3t\), and 12; their sum is \(t^2+7t+12\). Option A has factors 1 and 12, whose sum is 13, so its middle term would be \(13t\). Option C has the correct product 12 but gives \(2+6=8\), producing \(8t\). Option D gives both an incorrect product and an incorrect sum for this trinomial. The key check is always both product and sum, not just the constant term.