2026 AMC 10A Problem 24

1. Write the polynomial. The leading coefficient is $1$, so
$$f(x) = x^3 + bx^2 + cx + d,$$
where $b$, $c$, $d$ are integers from $-10$ to $10$. Each has $21$ possible values before we use the condition.

2. Name the two inputs. Let $u = 3 + \sqrt5$ and $v = 3 - \sqrt5$. Their sum and product are whole numbers:
$$u + v = 6, \qquad uv = 9 - 5 = 4.$$
Their difference is $u - v = 2\sqrt5$, which is not zero. The plan is to rewrite the condition using only $u + v$ and $uv$, so that the square roots disappear.

3. Subtract the two values. The condition $f(u) = f(v)$ is the same as $f(u) - f(v) = 0$:
$$f(u) - f(v) = (u^3 - v^3) + b(u^2 - v^2) + c(u - v).$$
The two $d$'s cancel. So $d$ does not matter at all; keep that in mind for the end.

4. Pull out the common factor $u - v$. Every piece contains a factor of $u - v$:
$$u^2 - v^2 = (u - v)(u + v), \qquad u^3 - v^3 = (u - v)(u^2 + uv + v^2).$$
(Check the second one by multiplying out: $(u - v)(u^2 + uv + v^2) = u^3 + u^2v + uv^2 - u^2v - uv^2 - v^3 = u^3 - v^3$.)

Why this helps: once $u - v$ is pulled out, what is left does not change if you swap $u$ and $v$. Expressions like that can always be written using just $u + v$ and $uv$, which are plain whole numbers here. So
$$f(u) - f(v) = (u - v)\left[(u^2 + uv + v^2) + b(u + v) + c\right].$$

5. Use the sum and product. Since $(u + v)^2 = u^2 + 2uv + v^2$,
$$u^2 + uv + v^2 = (u + v)^2 - uv = 36 - 4 = 32.$$
So
$$f(u) - f(v) = 2\sqrt5\,(32 + 6b + c).$$
The factor $2\sqrt5$ is not zero, so the condition holds exactly when
$$32 + 6b + c = 0, \qquad \text{that is,} \qquad c = -32 - 6b.$$

1. Find the allowed $b$. We need $c$ between $-10$ and $10$:

Larger or smaller values of $b$ push $c$ even further out, so there are exactly $4$ good pairs $(b, c)$.

7. Bring back $d$. Each good pair works with every one of the $21$ values of $d$. So the number of polynomials is
$$4 \times 21 = 84.$$

1. Check one. Take $b = -6$, $c = 4$, $d = 0$: $f(x) = x^3 - 6x^2 + 4x = x(x^2 - 6x + 4)$. Since $u + v = 6$ and $uv = 4$, both $u$ and $v$ are roots of $x^2 - 6x + 4$, so $f(u) = 0 = f(v)$. It works.

The answer is $\boxed{\textbf{(D)}\ 84}$.

The numbers $3 + \sqrt5$ and $3 - \sqrt5$ have a whole-number sum and product, and any expression that treats them the same way can be written with just that sum and product. Subtracting $f(v)$ from $f(u)$ and factoring out $u - v$ (using the difference of squares and the difference of cubes) leaves exactly such an expression. The same move works whenever a problem says $f(\text{one number}) = f(\text{another})$: subtract, factor out the difference, and see what is left.

Expand directly and compare the $\sqrt5$ parts.

1. $u^2 = (3 + \sqrt5)^2 = 9 + 6\sqrt5 + 5 = 14 + 6\sqrt5$.
2. $u^3 = u \cdot u^2 = (3 + \sqrt5)(14 + 6\sqrt5) = 42 + 18\sqrt5 + 14\sqrt5 + 30 = 72 + 32\sqrt5$.
3. So
$$f(u) = (72 + 14b + 3c + d) + (32 + 6b + c)\sqrt5.$$
4. For $v = 3 - \sqrt5$, every step is the same with $-\sqrt5$ in place of $\sqrt5$:
$$f(v) = (72 + 14b + 3c + d) - (32 + 6b + c)\sqrt5.$$
5. These are equal exactly when the $\sqrt5$ part is zero: $32 + 6b + c = 0$. This is the same condition as before, and the count is again $4 \times 21 = 84$.

Write $f(x) = x^3 + bx^2 + cx + d$ and let $u = 3 + \sqrt5$, $v = 3 - \sqrt5$, so $u + v = 6$, $uv = 4$ and $u - v = 2\sqrt5 \neq 0$. Then
$$f(u) - f(v) = (u - v)\left[u^2 + uv + v^2 + b(u + v) + c\right] = 2\sqrt5\left[(u+v)^2 - uv + 6b + c\right] = 2\sqrt5\,(32 + 6b + c).$$
So the condition is $c = -32 - 6b$, with $d$ unrestricted. The requirement $-10 \le c \le 10$ gives $-42 \le 6b \le -22$, so $b$ is one of $-7, -6, -5, -4$ (with $c = 10, 4, -2, -8$). With $21$ choices of $d$, there are $4 \cdot 21 = 84$ polynomials.

The answer is $\boxed{\textbf{(D)}\ 84}$.