2002 AMC 10B Problem 4

Before plugging in, look at the shape. The expression has three parts:
$$(3x - 2)(4x + 1)\ \ -\ \ (3x - 2)\cdot 4x\ \ +\ \ 1.$$
The last $+1$ stands alone; it is not inside either product.

1. The first two parts both contain $(3x - 2)$, so pull it out:
$$(3x-2)\big[(4x+1) - 4x\big] + 1.$$
2. Inside the bracket, $(4x + 1) - 4x = 1$.
3. So the whole thing is $(3x - 2)\cdot 1 + 1 = 3x - 1$.
4. At $x = 4$: $3 \cdot 4 - 1 = 11$.

The answer is $\boxed{\textbf{(D)}\ 11}$.

When two products share a factor, pulling it out usually makes most of the expression disappear. It also helps to read the expression carefully before doing anything: knowing that the final $+1$ is on its own is what keeps you from getting $0$.

Just plug in $x = 4$ right away. Then $3x - 2 = 10$, $4x + 1 = 17$ and $4x = 16$, so the value is
$$10 \cdot 17 - 10 \cdot 16 + 1 = 170 - 160 + 1 = 11.$$

The first two terms both contain $(3x - 2)$. Factor it out, being careful that the final $+1$ stands alone:
$$(3x-2)(4x+1) - (3x-2)\cdot 4x + 1 = (3x-2)\big[(4x+1) - 4x\big] + 1 = (3x-2)\cdot 1 + 1 = 3x - 1.$$

At $x = 4$ this is $3\cdot 4 - 1 = 11$.

The answer is $\boxed{\textbf{(D)}\ 11}$.