2023 AMC 10 A Problem 21

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To help you master 2023 AMC 10 A Problem 21, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

Related problems from past AMC exams, plus new practice problems that build the same skills

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2023 AMC10 A Problem 21

Solution

Solution 1

Factor Theorem and Degree Analysis

The given conditions provide four points on the graph of $P (x)$ . Since three of these points evaluate to $0$ , we can build the polynomial using its factors. We then use the fourth point to determine the minimal degree and the final unknown factor.

Step 1: Translate the root conditions into points on $P (x)$

If a value $r$ is a root of a function $f (x)$ , it means $f (r) = 0$ . We apply this definition to each given condition by substituting the given $x$ -value directly into the expression.

1 is a root of P (x) - 1 ⟹ P (1) - 1 = 0 ⟹ P (1) = 1

2 is a root of P (x - 2) ⟹ P (2 - 2) = 0 ⟹ P (0) = 0

3 is a root of P (3 x) ⟹ P (3 \cdot 3) = 0 ⟹ P (9) = 0

4 is a root of 4 P (x) ⟹ 4 P (4) = 0 ⟹ P (4) = 0

Step 2: Build the factored form of the polynomial

Because $P (0) = 0$ , $P (4) = 0$ , and $P (9) = 0$ , the Factor Theorem tells us that $x$ , $(x - 4)$ , and $(x - 9)$ must be factors of $P (x)$ . Therefore, $P (x)$ can be written as a product of these factors and another unknown polynomial $Q (x)$ .

P (x) = x (x - 4) (x - 9) \cdot Q (x)

Step 3: Determine the minimal degree of the polynomial

We want $P (x)$ to have the minimal possible degree and a leading coefficient of $1$ . First, test if $Q (x)$ can be a constant $c$ . To maintain a leading coefficient of $1$ , we must have $c = 1$ , giving $P (x) = x (x - 4) (x - 9)$ . Let’s check this against our remaining condition $P (1) = 1$ : $P (1) = 1 (1 - 4) (1 - 9) = 24$ . This contradicts $P (1) = 1$ , so $Q (x)$ cannot be a constant. Next, try making $Q (x)$ a polynomial of degree $1$ . To keep the overall leading coefficient $1$ , $Q (x)$ must also have a leading coefficient of $1$ . Thus, we can write $Q (x) = x - r$ , where $r$ is an unknown root.

P (x) = x (x - 4) (x - 9) (x - r)

Step 4: Solve for the unknown root

Use the condition $P (1) = 1$ to solve for $r$ .

P (1) = 1 (1 - 4) (1 - 9) (1 - r)

1 = 1 (- 3) (- 8) (1 - r)

1 = 24 (1 - r)

1 - r = \frac{1}{24}

r = 1 - \frac{1}{24} = \frac{23}{24}

Step 5: Identify the non-integer root and compute the final answer

The roots of $P (x)$ are the values of $x$ that make $P (x) = 0$ . Reading from our factored form, the roots are $0$ , $4$ , $9$ , and $\frac{23}{24}$ . The first three are integers. The single non-integer root is $\frac{23}{24}$ , which is already written as a fraction $\frac{m}{n}$ in simplest form (so $m$ and $n$ are relatively prime). We extract $m$ and $n$ and find their sum.

m = 23

n = 24

m + n = 23 + 24 = 47

Correct answer: D. $47$
The sum is

47

, which corresponds to choice D. Final Check: We can evaluate

P (x) = x (x - 4) (x - 9) (x - \frac{23}{24})

against the original conditions to ensure no mistakes were made. – Is

1

a root of

P (x) - 1

?

P (1) - 1 = 1 (- 3) (- 8) (1 - \frac{23}{24}) - 1 = 24 (\frac{1}{24}) - 1 = 0

. (True) – Is

2

a root of

P (x - 2)

?

P (2 - 2) = P (0) = 0

. (True) – Is

3

a root of

P (3 x)

?

P (3 \cdot 3) = P (9) = 0

. (True) – Is

4

a root of

4 P (x)

?

4 P (4) = 4 (0) = 0

. (True) All conditions are perfectly satisfied.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Definition of a Root of a Function:

If a value

r

is a root of a function

f (x)

, then

f (r) = 0

.

$f (x)$ can be any expression. $r$ must be substituted exactly for $x$ . The problem presents roots of modified functions like $P (3 x)$ . Substituting the root directly into the expression evaluates the function to $0$ , turning the word problem into clean algebraic equations.

Factor Theorem:

A polynomial $P (x)$ has a root at $x = r$ if and only if $(x - r)$ is a factor of $P (x)$ .

$P (r) = 0 ⟺ P (x) = (x - r) Q (x)$ for some polynomial $Q (x)$ . We discovered that $P (0) = 0$ , $P (4) = 0$ , and $P (9) = 0$ . The Factor Theorem allows us to immediately write $P (x)$ as a product involving $x$ , $(x - 4)$ , and $(x - 9)$ .

Leading Coefficient of a Factored Polynomial:

The leading coefficient of a product of polynomials is the product of their individual leading coefficients.

If $P (x) = A (x) \cdot B (x)$ , then $LC (P) = LC (A) \cdot LC (B)$ . Because $P (x) = x (x - 4) (x - 9) \cdot Q (x)$ is required to have a leading coefficient of $1$ , and the first three factors already have leading coefficients of $1$ , we can safely deduce that $Q (x)$ must also have a leading coefficient of $1$ .

Key Takeaway: Remember This Strategy for Next Time

A problem defines roots for transformations of a function, such as " $x = 3$ is a root of $P (3 x)$ ".

Try: Immediately substitute the given $x$ -value into the expression and set it equal to $0$ . Do not try to solve for $P (x)$ before evaluating.

Why: The definition of a root applies directly to the entire expression. Evaluating it strips away the confusing function transformations and provides concrete points on the original polynomial.

A polynomial is described by multiple roots (values where $P (x) = 0$ ).

Try: Use the Factor Theorem to set up a "skeleton" equation: $P (x) = (x - r_{1}) (x - r_{2}) \dots Q (x)$ .

Why: Factored form naturally enforces the root conditions without needing to solve large systems of equations for the coefficients of $a x^{n} + b x^{n - 1} + \dots$ .

A problem asks for a polynomial of "minimal degree".

Try: Assume the remaining unknown factor $Q (x)$ is a constant. If that contradicts another given condition, step up the degree by assuming $Q (x) = x - r$ .

Why: Polynomial degrees are integers. The "minimal degree" is simply the lowest integer degree that does not create a mathematical contradiction with the given points.

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2023 AMC 10 A Problem 21

In This Page:

2023 AMC10 A Problem 21