2002 AMC 10B Problems/Problem 4

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Problem

What is the value of

$(3x-2)(4x+1)-(3x-2)4x+1$

when $x=4$?

$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 10\qquad \mathrm{(D) \ } 11\qquad \mathrm{(E) \ } 12$

Solution

$(10)(17)-(10)(16)+1=170-160+1=11\Longrightarrow\mathrm{ {D} \ }$


One can also do most of the algebra, and only then substitute for $x$. Note that the first term $(3x-2)(4x+1)$ can be written as $(3x-2)4x + (3x-2)1$. Hence we get:

\begin{align*} & (3x-2)(4x+1)-(3x-2)4x+1 \\ &= (3x-2)4x + (3x-2)1 -(3x-2)4x+1 \\ &= 3x-2+1 \\ &= 3x-1 \\ &= 11. \end{align*}

See Also

2002 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions