Question #c7d51

Dec 12, 2016

$r \left(x\right) = \left({2}^{51} - 1\right) x + 2 - {2}^{51}$

Explanation:

${x}^{51} = {p}_{49} \left(x\right) \left({x}^{2} - 3 x + 2\right) + r \left(x\right)$

Here ${p}_{49} \left(x\right)$ represents a polynomial of degree $49$ and $r \left(x\right)$ is the division remainder. The remainder degree is always lower than the divisor. So ${p}_{49} \left(x\right)$ is the quotient, $\left({x}^{2} - 3 x + 2\right)$ is the divisor, and $r \left(x\right)$ the remainder. Note that

degree[${p}_{49} \left(x\right)$] = $49$
degree[$\left({x}^{2} - 3 x + 2\right)$] = $2$
degree[$r \left(x\right)$] = $1$

then

degree[${p}_{49} \left(x\right) \left({x}^{2} - 3 x + 2\right)$] = $51$

but $\left({x}^{2} - 3 x + 2\right) = \left(x - 1\right) \left(x - 2\right)$ and

the general $1$ degree polynomial representing $r \left(x\right)$ is $a x + b$ so

${x}^{51} = {p}_{49} \left(x\right) \left(x - 1\right) \left(x - 2\right) + a x + b$ The constants $a , b$ are obtained because we know two conditions to be obeyed.

$x = 1 \to {1}^{51} = a + b$ and
$x = 2 \to {2}^{51} = 2 a + b$

Solving for $a , b$ we get

$r \left(x\right) = \left({2}^{51} - 1\right) x + 2 - {2}^{51}$