2022 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 21 Hard

Let P(x) be a polynomial with rational coefficients such that when P(x) is divided by the polynomial x^{2}+x+1, the remainder is x+2, and when P(x) is divided by the polynomial x^{2}+1  the remainder is 2x+1. There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?

  • A.

    10

  • B.

    13

  • C.

    19

  • D.

    20

  • E.

    23

Answer:E

p(x) \div \left(x^2+x+1\right) R \ (x+2);

p(x) \div \left(x^2+1\right) R \ (2 x+1).

Let the remainder of p(x) \div\left(x^2+1\right)\left(x^2+x+1\right) be \theta(x)=A x^3+B x^2+cx+d.

Then \left\{\begin{array}{l}\theta(x)=\left(x^2+x+1\right)(m x+n)+(x+2) \\ \theta(x)=\left(x^2+1\right)(k x+b)+(2 x+1)\end{array}\right..

Simplify to get

\left\{\begin{array}{l}\theta(x)=m x^3+(n+m) x^2+(m+n-1) x+n+2 \\ \theta(x)=k x^3+b x^2+(k+2) x+b+1\end{array}\right.

So \left\{\begin{array}{l}m=k \\ m+n=b \\ m+n+1=k+2 \\ n+2=b+1\end{array}\right.

and \left\{\begin{array}{l}k=1 \\ m=1 \\ n=1 \\ b=2\end{array}\right..

\theta(x)=x^3+2 x^2+3 x+3,

Answer is 1^2+2^2+3^2+3^2=23.

Related Problems
Problem 19 2022 AMC 10 B
Problem 20 2022 AMC 10 B
Problem 22 2022 AMC 10 B
Problem 23 2022 AMC 10 B
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