Finding the remainder

Let $p(x) = x+x^3+x^9+x^{27}+x^{81}+x^{243}$ ; then:

$\begin{aligned} p(x) & = (x^2-1)q(x) + \color{#3D99F6}r(x) & \small \color{#3D99F6} \text{where }q(x) = \text{ quotient, }r(x) = \text{ remainder.} \\ & = (x+1)(x-1)q(x) + \color{#3D99F6} ax + b & \small \color{#3D99F6} r(x) \text{ is 1 degree lesser than divisor }x^2-1 \\ p(-1) & = 0 - a + b \\ \implies -a + b & = -6 & \color{#D61F06} ...(1) \\ p(1) & = 0 + a + b \\ \implies a + b & = 6 & \color{#D61F06} ...(2) \\ (2)-(1): \quad a & = 6 \\ (1)+(2): \quad b & = 0 \end{aligned}$

$\implies a+b = 6+0 = \boxed{6}$

We have that $x + x^3 + x^9 + x^{27} + x^{81} + x^{243} = (x^2 - 1) q(x) + ax + b$ for some polynomial $q(x)$ . Setting $x = 1$ , we get $a + b = 6$ .

Let us denote by $q(x)$ the quotient resulting from the division of the given polynomial by $x^2-1$ , and let $ax+b$ be the remainder. Note that the division of a polynomial by a quadratic trinomial leaves a remainder which is a binomial of the first degree. Thus

$x+x^3+x^9+x^{27}+x^{81}+x^{243}=q(x)(x^2-1)+ax+b$ $\color{#D61F06}(1)$

Since, the divisor is $x^2-1$ , we have that $x=±\sqrt1=±1$ .

On putting $x=1$ in $\color{#D61F06}(1)$ , we obtain

$6=a+b$

On putting $x=-1$ in $\color{#D61F06}(1)$ , we obtain

$-6=-a+b$

Whence,

$b=0$ and $a=6$

Thus the remainder is $6x$ .

Finally,

$a+b=6+0=$ $\boxed{\color{#3D99F6}6}$