Context: The expression $-16(5x-3)^2 + 4(5x-4)^2$ can be rewritten as $\dfrac{a}{6}x^2 + \dfrac{b}{6}x + \dfrac{c}{6}$ for constants $a$, $b$, $c$. We need $a + b + c$.

Answer: A) $-360$

Why: Expand: $-16(25x^2 - 30x + 9) = -400x^2 + 480x - 144$ and $4(25x^2 - 40x + 16) = 100x^2 - 160x + 64$. The sum is $-300x^2 + 320x - 80$. Matching coefficients, $\dfrac{a}{6} = -300$, $\dfrac{b}{6} = 320$, $\dfrac{c}{6} = -80$, so $a = -1800$, $b = 1920$, $c = -480$ and $a + b + c = -360$. Shortcut: $\dfrac{a + b + c}{6}$ is the expression's value at $x = 1$, which is $-16(2)^2 + 4(1)^2 = -60$, so $a + b + c = 6(-60) = -360$.

Traps:

• B) Arithmetic slip; $-180$ would be $3(-60)$, but the sum must be $6$ times the value at $x = 1$.
• C) $-60$ is $-300 + 320 - 80$, the sum of the simplified coefficients; it forgets that $a$, $b$, $c$ are $6$ times those numbers.
• D) Divides by $6$ instead of multiplying: $-60 \div 6 = -10$.