Choose the correct factorised form of (p^2-p-20).

Correct answer: (p+4)(p-5)

The direct answer is A: $(p+4)(p-5)$. For a monic quadratic $p^2+bp+c$, find two numbers whose product is c and whose sum is b. Here the product must be -20 and the sum must be -1. The numbers 4 and -5 satisfy both conditions: $4×(-5)=-20$ and $4+(-5)=-1$. Therefore $p^2-p-20=(p+4)(p-5)$. Expanding confirms it: $p^2-5p+4p-20=p^2-p-20$. B has factors -4 and 5; their product is -20 but their sum is 1, giving $p^2+p-20$. C uses -2 and 10, whose sum is 8, so it gives $p^2+8p-20$. D uses 20 and -1, whose sum is 19, producing $p^2+19p-20$. Always check both the product and the sum, not just the constant term.