Question

Let $P(x) = x^{2025}$ and $Q(x) = x^4 + x^3 + 2x^2 + x + 1$. Let $R(x)$ be the polynomial remainder when the polynomial $P(x)$ is divided by the polynomial $Q(x)$. Find $R(3)$.

• A. 21
• B. 53
• C. 7
• D. 1

Answer 1

Answer: (B)

53

Answer 2

We are given $P(x) = x^{2025}$ and $Q(x) = x^4 + x^3 + 2x^2 + x + 1$. We need to find the remainder $R(x)$ when $P(x)$ is divided by $Q(x)$, and then evaluate $R(3)$.

First, let's factorize $Q(x)$: $Q(x) = x^4 + x^3 + 2x^2 + x + 1$ We can group terms: $Q(x) = x^2(x^2 + x + 1) + (x^2 + x + 1)$ $Q(x) = (x^2 + 1)(x^2 + x + 1)$

The roots of $Q(x)$ are the roots of $x^2+1=0$ and $x^2+x+1=0$.

Case 1: Roots of $x^2+x+1=0$. Let $\omega$ be a root. Then $\omega^2+\omega+1=0$. Multiplying by $(\omega-1)$, we get $\omega^3-1=0$, which means $\omega^3=1$. If $x$ is a root of $Q(x)$, then $P(x) = R(x)$. So, $R(x) = x^{2025}$. Since $\omega^3=1$, we can reduce the exponent $2025$ modulo $3$. $2025 \div 3 = 675$ with a remainder of $0$. So, $\omega^{2025} = (\omega^3)^{675} = 1^{675} = 1$. Thus, for any root $x$ of $x^2+x+1=0$, $R(x)=1$. This implies $R(x) \equiv 1 \pmod{x^2+x+1}$.

Case 2: Roots of $x^2+1=0$. Let $i$ be a root. Then $i^2+1=0$, so $i^2=-1$. The powers of $i$ cycle with a period of 4: $i^1=i, i^2=-1, i^3=-i, i^4=1$. We need to reduce the exponent $2025$ modulo $4$. $2025 \div 4 = 506$ with a remainder of $1$. So, $i^{2025} = i^{4 \times 506 + 1} = (i^4)^{506} \cdot i^1 = 1^{506} \cdot i = i$. Thus, for any root $x$ of $x^2+1=0$, $R(x)=x$. This implies $R(x) \equiv x \pmod{x^2+1}$.

Now we have two congruences for $R(x)$:

1. $R(x) \equiv 1 \pmod{x^2+x+1}$
2. $R(x) \equiv x \pmod{x^2+1}$

Since $\deg(Q(x))=4$, the remainder $R(x)$ has a degree of at most 3. Let $R(x) = ax^3+bx^2+cx+d$.

From $R(x) \equiv 1 \pmod{x^2+x+1}$, we can write $R(x) = A(x)(x^2+x+1) + 1$. Since $\deg(R(x)) \le 3$, $A(x)$ must be a linear polynomial, say $mx+n$. $R(x) = (mx+n)(x^2+x+1) + 1 = mx^3 + mx^2 + mx + nx^2 + nx + n + 1 = mx^3 + (m+n)x^2 + (m+n)x + (n+1)$.

From $R(x) \equiv x \pmod{x^2+1}$, we can write $R(x) = B(x)(x^2+1) + x$. Since $\deg(R(x)) \le 3$, $B(x)$ must be a linear polynomial, say $px+q$. $R(x) = (px+q)(x^2+1) + x = px^3 + px + qx^2 + q + x = px^3 + qx^2 + (p+1)x + q$.

Equating the two expressions for $R(x)$: $mx^3 + (m+n)x^2 + (m+n)x + (n+1) = px^3 + qx^2 + (p+1)x + q$

Comparing coefficients:

1. $m = p$
2. $m+n = q$
3. $m+n = p+1$
4. $n+1 = q$

From (2) and (4), we have $m+n = n+1$, which implies $m=1$. Since $m=p$ from (1), we have $p=1$. Substitute $m=1$ into (2): $1+n = q$. This is consistent with (4). Substitute $m=1$ into (3): $1+n = p+1$. Since $p=1$, this becomes $1+n = 1+1$, so $1+n = 2$, which gives $n=1$. Now we find $q$ using $q=n+1$: $q=1+1=2$.

So, $m=1, n=1, p=1, q=2$.

Substitute these values back into the expressions for $R(x)$: Using the first form: $R(x) = 1x^3 + (1+1)x^2 + (1+1)x + (1+1) = x^3 + 2x^2 + 2x + 2$. Using the second form: $R(x) = 1x^3 + 2x^2 + (1+1)x + 2 = x^3 + 2x^2 + 2x + 2$. Both forms yield the same remainder $R(x) = x^3 + 2x^2 + 2x + 2$.

Finally, we need to find $R(3)$: $R(3) = (3)^3 + 2(3)^2 + 2(3) + 2$ $R(3) = 27 + 2(9) + 6 + 2$ $R(3) = 27 + 18 + 6 + 2$ $R(3) = 53$.

The first part of the question regarding $f(n)$ and $T$ is incomplete and cannot be solved. The number 7 is likely an answer to that part. We have solved problem 21.