2003 AMC 10A Problem 5

The expression $(d-1)(e-1)$ has a "$-1$" in each factor, which suggests checking whether $1$ itself is a solution.

1. Put $x = 1$ into $2x^2 + 3x - 5$:
$$2(1)^2 + 3(1) - 5 = 2 + 3 - 5 = 0.$$
2. It gives $0$, so $x = 1$ is a solution. That means one of $d$ and $e$ is $1$.
3. Say $d = 1$. Then the first factor is $d - 1 = 0$.
4. Zero times anything is zero, so $(d-1)(e-1) = 0$, whatever $e$ is.

(If you want to see the other solution: $2x^2 + 3x - 5 = (x - 1)(2x + 5)$, so $e = -\frac{5}{2}$, and $(1 - 1)\left(-\frac{5}{2} - 1\right) = 0$.)

The answer is $\boxed{\textbf{(B)}\ 0}$.

If the numbers in front of the terms add to zero (here $2 + 3 - 5 = 0$), then $x = 1$ is always a solution, because putting in $1$ just adds those numbers up. More generally, whenever a question asks for $(d - k)(e - k)$ for the two solutions, try putting $x = k$ into the equation; it is often faster than solving.

Use the sum and product of the solutions. For $2x^2 + 3x - 5 = 0$, the sum is $d + e = -\frac{3}{2}$ and the product is $de = -\frac{5}{2}$. Multiplying out,
$$(d-1)(e-1) = de - d - e + 1 = de - (d + e) + 1 = -\frac{5}{2} + \frac{3}{2} + 1 = 0.$$