2014 AMC 10B Problem 20

1. Spot the pattern. Only $x^4$ and $x^2$ appear, and $x^4 = (x^2)^2$. So think of $x^2$ as one number, say $u$. The expression is $u^2 - 51u + 50$.
2. Factor. We need two numbers that multiply to $50$ and add to $51$: they are $1$ and $50$. So
$$x^4 - 51x^2 + 50 = (x^2 - 1)(x^2 - 50).$$
3. When is a product negative? When one factor is positive and the other is negative. The factor $x^2 - 1$ is always the larger of the two, so it must be the positive one:
$$x^2 - 1 > 0 \quad\text{and}\quad x^2 - 50 < 0, \qquad\text{that is,}\qquad 1 < x^2 < 50.$$
4. List the squares between $1$ and $50$ (not including $1$): $4, 9, 16, 25, 36, 49$. These come from $x = 2, 3, 4, 5, 6, 7$. ($8^2 = 64$ is too big.)
5. Each of those also works as a negative number, since $(-x)^2 = x^2$. That doubles the count: $6 \times 2 = 12$.
6. Check the edges. At $x = 7$: $(49 - 1)(49 - 50) = 48 \times (-1) = -48$, negative. At $x = 1$: $0 \times (-49) = 0$, not negative. At $x = 8$: $63 \times 14$, positive.

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

An expression with only even powers of $x$ is really a quadratic in $x^2$, so factor it the usual way. That turns a hard sign question about $x$ into a simple range for $x^2$. Remember that each positive value of $x^2$ comes from two integers, $x$ and $-x$.

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 $\boxed{\textbf{(C)}\ 12}$.