AMC 10 Step by Step

2002 AMC 10B Problem 4

P1-10~1 minPrint

What is the value of (3x−2)(4x+1)−(3x−2)4x+1(3x - 2)(4x + 1) - (3x - 2)4x + 1 when x=4x=4 ?

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Key insight

The first two parts both contain (3x - 2). Pull it out and the rest cancels, leaving 3x - 1, which is 11 when x = 4.

Solution

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

1. The first two parts both contain (3x−2)(3x - 2), so pull it out:
(3x−2)[(4x+1)−4x]+1.(3x-2)\big[(4x+1) - 4x\big] + 1.
2. Inside the bracket, (4x+1)−4x=1(4x + 1) - 4x = 1.
3. So the whole thing is (3x−2)⋅1+1=3x−1(3x - 2)\cdot 1 + 1 = 3x - 1.
4. At x=4x = 4: 3⋅4−1=113 \cdot 4 - 1 = 11.

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

Why this works

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+1 is on its own is what keeps you from getting 00.

Alternative approach

Just plug in x=4x = 4 right away. Then 3x−2=103x - 2 = 10, 4x+1=174x + 1 = 17 and 4x=164x = 16, so the value is
10⋅17−10⋅16+1=170−160+1=11.10 \cdot 17 - 10 \cdot 16 + 1 = 170 - 160 + 1 = 11.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The first two terms both contain (3x−2)(3x - 2). Factor it out, being careful that the final +1+1 stands alone:
(3x−2)(4x+1)−(3x−2)⋅4x+1=(3x−2)[(4x+1)−4x]+1=(3x−2)⋅1+1=3x−1.(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=4x = 4 this is 3⋅4−1=113\cdot 4 - 1 = 11.

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

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site. Also appeared as 2002 12B #2.