Find the remainder when $x^{100}$ is divided by $x^2-3x+2$

Question

We have to find the remainder when $x^{100}$ is divided by $x^2-3x+2$.I tried to use the remainder theorem but am not just able to solve it.please help.

Answer 1

HINT:

$$x^{100}=(x^2-3x+2)Q(x)+ax+b$$

$$x^{100}=(x-1)(x-2)Q(x)+ax+b$$

Apply roots of $(x^2-3x+2)$ to find $a,b$

Answer 2

Let $p(x) = x^{100}$ and $q(x) = x^2 – 3x + 2 = x^2 – 2x – x + 2 = x(x– 2) – 1(x – 2) = (x – 2)(x – 1)$

Let $Q(x)$ and $r(x)$ be the quotient and remainder when $p(x)$ is divided by $q(x)$. Hence by division algorithm, we have $p(x) = q(x)\cdot Q(x) + r(x)$

That is, $x^{100} = (x – 2)(x – 1)Q(x) + r(x)$

Let $r(x) = (ax + b)$ where $0 < \deg r(x) < 2$

$x^{100} = (x – 2)(x – 1)Q(x) + (ax + b)\qquad (1)$

Put $x = 1$ in Equation $(1)$, we get $1 = 0 + (a + b)$

Hence $(a + b) = 1\qquad(2)$

Now put x = 2 in equation (1), we get

$2^{100} = 0 + (2a + b)$

$2a + b = 2^{100}\qquad(3)$

Solving $(2)$ and $(3)$, we get $a = (2^{100} – 1)$ and $b = 2 – 2^{100}$

Therefore the remainder $= (ax + b) = (2^{100} – 1)x + (2 – 2^{100})$