Remainder Problem Help

Let P(x)=x3+x+1 and Q(x)=x3−x+1.

We are given that when f(x) is divided by P(x), the remainder is 3, so we can write f(x)=P(x)q1​(x)+3 for some polynomial q1​(x).

When f(x) is divided by Q(x), the remainder is x+1, so we can write f(x)=Q(x)q2​(x)+x+1 for some polynomial q2​(x).

We want to find the remainder when f(x) is divided by x6+2x3−x2+1.

Note that

$$\begin{align*} P(x) Q(x) &= (x^3 + x + 1)(x^3 - x + 1) \ &= (x^3 + 1)^2 - x^2 \ &= x^6 + 2x^3 + 1 - x^2 \ &= x^6 + 2x^3 - x^2 + 1. \end{align*}$$

Let R(x)=x6+2x3−x2+1. Then R(x)=P(x)Q(x).

From f(x)=P(x)q1​(x)+3,

$$\begin{align*} f(x) &= P(x) q_1(x) + 3 \ &= P(x) [Q(x) q_3(x) + r_1(x)] + 3 \ &= R(x) q_3(x) + P(x) r_1(x) + 3 \end{align*}$$

for some polynomials q3​(x) and r1​(x), where degr1​(x)<3.

From f(x)=Q(x)q2​(x)+x+1,

$$\begin{align*} f(x) &= Q(x) q_2(x) + x+1 \ &= Q(x) [P(x) q_4(x) + r_2(x)] + x+1 \ &= R(x) q_4(x) + Q(x) r_2(x) + x+1 \end{align*}$$

for some polynomials q4​(x) and r2​(x), where degr2​(x)<3.

Since f(x)≡3(modP(x)) and f(x)≡x+1(modQ(x)), we must have r(x)≡3(modP(x)) and r(x)≡x+1(modQ(x)).

If r(x)=ax2+bx+c, then

$$\begin{align*} a(x^2+x+1) + b(x+1) + c &\equiv 3 \pmod{x^3+x+1} \ a(x^2-x+1) + b(x+1) + c &\equiv x+1 \pmod{x^3-x+1} \end{align*}$$

Since r(x) must have degree less than 6, let r(x)=ax5+bx4+cx3+dx2+ex+f.

Since r(x)≡3(modx3+x+1) and r(x)≡x+1(modx3−x+1), we can let r(x)=3.

Then r(2)=3.

Final Answer: The final answer is 3​