What are the factors of \(t^2+t-6\)?

Correct answer: \((t+3)(t-2)\)

The direct answer is A: \((t+3)(t-2)\). For \(t^2+bt+c\), find two numbers whose product is the constant term and whose sum is the coefficient of t. Here the product must be -6 and the sum must be 1. The numbers 3 and -2 satisfy both conditions: \(3(-2)=-6\) and \(3+(-2)=1\). Hence \(t^2+t-6=(t+3)(t-2)\). A expands to \(t^2-2t+3t-6=t^2+t-6\). B uses -3 and 2, whose sum is -1, so it gives \(t^2-t-6\). C uses 1 and -6, whose product is -6 but sum is -5. D uses 6 and -1, whose sum is 5. Check both sum and product, not just one.