2022 AMC 10 B Problem 21

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To help you master 2022 AMC 10 B 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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2022 AMC10 B Problem 21

Solution

Solution 1

Undetermined Coefficients Method

Because the lowest degree polynomial satisfying both conditions has degree at most 3, we can express $P(x)$ using the polynomial division algorithm for each divisor. Expanding both expressions and equating their coefficients will reveal the unknown polynomial.

Polynomial Division Algorithm: Any polynomial $P(x)$ can be written as $P(x)=Q(x)D(x)+R(x)$, where $D(x)$ is the divisor, $Q(x)$ is the quotient, and $R(x)$ is the remainder with degree less than that of $D(x)$.

Step 1: Determine the degree and set up the division algorithm

Since the divisors $x^2+x+1$ and $x^2+1$ have no common roots, the polynomial of least degree satisfying both remainder conditions has a degree up to $2+2-1=3$. The remainder for each quadratic divisor is linear, so the quotient must be linear.

$$P(x)=(a_{1}x+b_1)(x^2+x+1)+x+2$$

$$P(x)=(a_{2}x+b_2)(x^2+1)+2x+1$$

Step 2: Standardize the leading coefficient and expand

Both forms must represent the exact same polynomial $P(x)$, which means their leading coefficient $x^3$ must match. Thus, $a_1=a_2=a$. We let $b_1=k$ and $b_2=m$, and expand both expressions.

$$P(x)=(ax+k)(x^2+x+1)+x+2=a{x}^3+(a+k)x^2+(a+k+1)x+k+2$$

$$P(x)=(ax+m)(x^2+1)+2x+1=a{x}^3+m{x}^2+(a+2)x+m+1$$

Step 3: Equate the corresponding coefficients

Because the two expanded forms represent the same polynomial, the coefficients of $x^2$, $x$, and the constant terms must respectively be equal.

$$a+k=m$$

$$a+k+1=a+2$$

$$k+2=m+1$$

Step 4: Solve the resulting system of equations

We can solve for $k$, $m$, and $a$ directly from the system.

$$a+k+1=a+2⟹k=1$$

$$1+2=m+1⟹m=2$$

$$a+1=2⟹a=1$$

Step 5: Construct $P(x)$ and calculate the sum of the squares of its coefficients

Substitute $a=1$ and $k=1$ back into the first expression to find $P(x)$. Finally, identify the coefficients and find the sum of their squares.

$$P(x)=(x+1)(x^2+x+1)+x+2$$

$$P(x)=(x^3+2x^2+2x+1)+x+2=x^3+2x^2+3x+3$$

$$1^2+2^2+3^2+3^2=1+4+9+9=23$$

Correct answer: E. $23$
The sum of the squares of the coefficients is 23 (Choice E). Check with the second divisor form: substituting $a=1$ and $m=2$ gives $P(x)=(x+2)(x^2+1)+2x+1=x^3+2x^2+x+2+2x+1=x^3+2x^2+3x+3$. Both forms yield the same polynomial.

Solution 2

Polynomial Modular Arithmetic Method

Instead of expanding the full polynomial and solving a system, we can satisfy the first condition by defining $P(x)=A(x)(x^2+x+1)+x+2$, and then substitute this form into the second condition using polynomial congruences.

Polynomial Congruences: Working modulo a polynomial $D(x)$ allows us to replace $D(x)$ with 0 in any expression. For example, working modulo $x^2+1$ means we can replace any instance of $x^2$ with $-1$.

Step 1: Express $P(x)$ using the first condition

By the division algorithm, $P(x)$ leaves a remainder of $x+2$ when divided by $x^2+x+1$. We know $P(x)$ has a maximum degree of 3, so the quotient $A(x)$ must be a linear polynomial.

$$P(x)=A(x)(x^2+x+1)+x+2$$

Step 2: Apply the second condition using modular arithmetic

We are given that $P(x)$ leaves a remainder of $2x+1$ when divided by $x^2+1$. This translates to the congruence $P(x)\equiv 2x+1(\mathrm{mod}{x}^2+1)$. We substitute our expression for $P(x)$ into this congruence.

$$A(x)(x^2+x+1)+x+2\equiv 2x+1(\mathrm{mod}{x}^2+1)$$

Step 3: Simplify the congruence using $x^2\equiv -1$

Working modulo $x^2+1$, we can substitute $-1$ wherever we see $x^2$. We simplify $x^2+x+1$ and subtract $x+2$ from both sides of the congruence.

$$x^2+x+1\equiv -1+x+1\equiv x(\mathrm{mod}{x}^2+1)$$

$$x\cdot A(x)+x+2\equiv 2x+1(\mathrm{mod}{x}^2+1)$$

$$x\cdot A(x)\equiv x-1(\mathrm{mod}{x}^2+1)$$

Step 4: Solve for the quotient polynomial $A(x)$

To isolate $A(x)$, we multiply both sides by $-x$. This is a clever choice because $(-x)(x)=-x^2$, and since $x^2\equiv -1$, we have $-x^2\equiv 1(\mathrm{mod}{x}^2+1)$.

$$(-x)\cdot x\cdot A(x)\equiv (-x)(x-1)(\mathrm{mod}{x}^2+1)$$

$$-x^{2}A(x)\equiv -x^2+x(\mathrm{mod}{x}^2+1)$$

$$1\cdot A(x)\equiv 1+x(\mathrm{mod}{x}^2+1)$$

Step 5: Find $P(x)$ and calculate the sum of squares

Since we want the polynomial of least degree, we choose $A(x)=x+1$. We substitute $A(x)$ back into our original formula for $P(x)$ and sum the squares of its coefficients.

$$P(x)=(x+1)(x^2+x+1)+x+2=x^3+2x^2+3x+3$$

$$1^2+2^2+3^2+3^2=23$$

Correct answer: E. $23$
The sum of the squares is 23 (Choice E). The logic guarantees the minimal degree polynomial directly without ever requiring a system of equations for the coefficients.

Solution 3

Complex Root Substitution Method

The remainder of a polynomial divided by a quadratic can be isolated by substituting the complex roots of the quadratic into the polynomial. Evaluating $P(x)$ at these roots produces a system of equations matching real and imaginary parts.

Generalized Polynomial Remainder Theorem: If $P(x)$ leaves a remainder of $R(x)$ when divided by $D(x)$, evaluating $P(x)$ at any root $r$ of $D(x)=0$ gives $P(r)=R(r)$.

Step 1: Define $P(x)$ and substitute a root of $x^2+1=0$

Let $P(x)=a{x}^3+b{x}^2+cx+d$. The roots of $x^2+1=0$ are $i$ and $-i$. The remainder when dividing by $x^2+1$ is $2x+1$, so $P(i)=2i+1$. We evaluate $P(i)$ and equate the real and imaginary parts.

$$P(i)=a(i^3)+b(i^2)+c(i)+d=-ai-b+ci+d$$

$$P(i)=(d-b)+(c-a)i$$

$$d-b=1$$

$$c-a=2$$

Step 2: Substitute a root of $x^2+x+1=0$

Let $\omega$ be a complex root of $x^2+x+1=0$. This implies ${\omega }^2=-\omega -1$ and ${\omega }^3=1$. The remainder when dividing by $x^2+x+1$ is $x+2$, so $P(\omega )=\omega +2$. Because the coefficients $a,b,c,d$ are rational, we can equate the constant parts and the $\omega$ parts just as we did with imaginary numbers.

$$P(\omega )=a({\omega }^3)+b({\omega }^2)+c(\omega )+d$$

$$P(\omega )=a(1)+b(-\omega -1)+c(\omega )+d=(a-b+d)+(c-b)\omega$$

$$c-b=1$$

$$a-b+d=2$$

Step 3: Solve the system of four equations

We now have a clean system of four linear equations for the four coefficients. We solve them systematically.

$$d-b=1⟹a+(d-b)=2⟹a+1=2⟹a=1$$

$$c-a=2⟹c=1+2=3$$

$$c-b=1⟹b=3-1=2$$

$$d-b=1⟹d=2+1=3$$

Step 4: Calculate the final answer

The polynomial is $P(x)=1x^3+2x^2+3x+3$. We find the sum of the squares of these coefficients.

$$1^2+2^2+3^2+3^2=1+4+9+9=23$$

Correct answer: E. $23$
The sum of the squares is 23 (Choice E). The complex root substitution directly extracts the coefficients without polynomial long division.

Similar Problems in Real Exams

The following five AMC 10 B problems are closely related to AMC 10 B 2022 Question 21. These real AMC 10 B problems test the same key concept—Algebra. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2025 AMC 10 B Problem 10

2022 AMC 10 B Problem 7

2021 AMC 10 B Problem 15

2025 AMC 10 B Problem 1

2025 AMC 10 B Problem 3

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Chinese Remainder Theorem for Polynomials:

Applies when $D_1(x)$ and $D_2(x)$ are relatively prime. The polynomial of least degree satisfying both has a degree strictly less than $\mathrm{deg}(D_1)+\mathrm{deg}(D_2)$. It guarantees that the minimal degree of $P(x)$ in this problem is at most $2+2-1=3$, allowing us to confidently assume $P(x)$ is a cubic polynomial.

• Polynomial Division Algorithm:

$P(x)$ is the dividend, $D(x)$ is the divisor, $Q(x)$ is the quotient, and $R(x)$ is the remainder. The degree of $R(x)$ must be less than the degree of $D(x)$. It forms the foundation of all three methods, allowing us to explicitly write out the structure of $P(x)$ given its remainders.

• Generalized Polynomial Remainder Theorem:

Applies to any complex root $r$ of the divisor $D(x)$. By substituting roots like $x=i$ or $x=\omega$, we bypass the unknown quotient entirely and create direct linear equations for the coefficients.

• Complex Cube Roots of Unity:

Used whenever evaluating expressions involving the polynomial $x^2+x+1$. It radically simplifies the substitution in Method 3, letting us reduce cubic and quadratic powers of $\omega$ to linear terms.

Key Takeaway: Remember This Strategy for Next Time

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