2014 AMC 10B Problem 20

Let $u = x^2$. Then the expression is $u^2 - 51u + 50$, and since $1 \cdot 50 = 50$ and $1 + 50 = 51$ it factors:
$$x^4 - 51x^2 + 50 = (x^2 - 1)(x^2 - 50).$$

The product of two factors is negative when they have opposite signs. Because $x^2 - 1 > x^2 - 50$ always, that means
$$x^2 - 1 > 0 > x^2 - 50 \quad\Longleftrightarrow\quad 1 < x^2 < 50.$$

The perfect squares strictly between $1$ and $50$ are $4, 9, 16, 25, 36, 49$, so $|x| \in \{2, 3, 4, 5, 6, 7\}$. Each of these six absolute values gives two integers, $x$ and $-x$, for a total of $12$.

The answer is $\boxed{\textbf{(C)}\ 12}$.

A polynomial in even powers only is a quadratic in $x^2$, and factoring it turns a sign question into an interval for $x^2$. Once the inequality is on $x^2$, remember that every positive value of $x^2$ corresponds to two integers.

Check the boundary: at $x = \pm1$ and $x = \pm 7$ the expression is $0$ and $2401 - 2499 + 50 = -48$ respectively, so $7$ is in and $1$ is out; at $x = 8$ it is $4096 - 3264 + 50 > 0$. The integers from $-7$ to $7$ excluding $-1, 0, 1$ number $15 - 3 = 12$.

• Counting only positive x (six values) and forgetting each has a negative twin, or including x = 1 and x = 7 endpoints carelessly.
• Including $x = 0$, where the expression equals $50 > 0$, or $x = \pm 1$, where it equals $0$ (not negative), giving $14$ or $16$.