When the polynomials $p(x)$ and $q(x)$ are divided by $x^2+2$ the rests are $ax+b$ respectively $cx+d$. What are the rest when $p(x)+q(x)$ and $p(x)q(x)$ divided by $x^2+2$?
I do not know where to start since I can't use the remainder theorem. Right?
The hint:
There are polynomials $p_1$ and $q_1$ for which $$p(x)+g(x)=p_1(x)(x^2+2)+q_1(x)(x^2+2)+(a+c)x+b+d.$$ Thus, the first answer is $(a+c)x+b+d.$
For the second use the similar reasoning, but you need to work also with degree.
Let $p(x)=(x^2+2)u(x)+ax+b$ and $q(x)=(x^2+2)v(x)+cx+d$ , then: $$p(x)+q(x)=(x^2+2)(u(x)+v(x))+ax+b+cx+d$$ So the rest of the division is: $$(a+b)x+b+d$$
Now: $$p(x)q(x)=(x^2+2)^2u(x)v(x)+(x^2+2)(ax+b+cx+d)+(ax+b)(cx+d)$$ so to get the final result we have to evaluate $(ax+b)(cx+d)=l(x^2+2)+mx+n$. It is easy to see that $l=ac$ so the final rest is $$mx+n=(ad+bc)x+bd-2ac$$
let r(x) s(x) is polynomial. we can write p(x)=(x^2+2)r(x)+ax+b q(x)=(x^2+2)s(x)+cx+d
let T(x) is polynomial. so we get p(x)+q(x)=(x^2+2x)T(x)+(a+c)x+b+d so rest is (a+c)x+b+d (cuz (a+c)x+b+d is first order polynomial)
considering p(x)q(x)’s rest , we just do same thing like what we do in p(x)+q(x) ,and get the rest
For the first question the remainders add up to $ax+b+cx+d$
For the second question the remainders multiply to $$(ax+b)(cx+d)$$ divided by $x^2+2$
That is $(ad+ bc-2ac)x+bc$
$$\begin{align} p \,&=\qquad a\ x\ +\ b\,\ +\,\ (x^2+2)\,f \\ q \,&=\qquad c\ x\ +\ d\ \, +\, \ (x^2+2)\,g\\[.2em] \hline \Rightarrow\ \ p+q &= (a\!+\!c)x + b\!+\!d + (x^2+2)(f+g) \end{align}$$
Multiplication is similar but we need to reduce the remainder product $\!\bmod x^2\!+\!2\,$ via $\,x^2\equiv -2,\,$ or, equivalently use $\,x^2\, = -2 + (x^2+2)$ to eliminate the $\,x^2\,$ term.
This is better done more conceptually using the Congruence Sum & Product Rules.
