Question

find the remainder when x51 is divided by x2- 3x + 2​

Answer 1

Firstly, the actual question is,

(x^51) ÷ (x^2-3x+2)

The solution is given as follows.

Let's consider that,

p(x) = x ^51

and, q(x) = x ^2 – 3x + 2 = (x – 1) (x – 2)

Now, when p(x) is divided by q(x), then as per the division algorithm there exists Q(x) and R(x) = ax+b , such that  :-

x^51 = Q(x)(x-1)(x-2) + ax +b [where Q(x) is the quotient & ax+b is the remainder ]

For (x = 1) we get,

1^51 = Q(1)(1-1)(1-2) + a+b [then a+b = 1 ]

and

2^51 = Q(2)(2-1)(2-2) + 2a + b [then 2a+b = 2^51 ]

Now, we solve :-

a+b = 1

2a+b = 2^51 [after solving this equation ]

a = 2^(51) - 1

b = 2- 2^(51)

Ans) The remainder will be x(2^51 - 1) + 2-2^51

Answer 2

The answer is $r(x)=(2^{51}-1)x+(2-2^{51})$

Step-by-step explanation:

Given,

Find the remainder when $x^{51}$ is divided by $x^{2} -3x+2$

Let,

$f(x)=x^{51}$ and $g(x)=x^{2} -3x+2$

We can now say,

$f(x)=(g(x)\times q(x))+r(x)$

Where $0\leq deg(r(x))<deg(g(x))$

Since $g(x)$ is quadratic,

$r(x)=ax+b$

∴ $x^{51}=(x-1)(x-2)q(x)+(ax+b)$

What if you put,

$x=1 and 2$

Let's for $x=1$,

$1^{51}=(1-1)(1-2)q(1)+(a+b)$

⇒$a+b=1$__1

For $x=2$,

$2^{51}=(2-1)(2-2)q(2)+(2a+b)$

⇒$2a+b=2^{51}$__2

By solving equation-1 and 2,

$a=2^{51}-1$ and $b=2-2^{51}$

So, $r(x)=(2^{51}-1)x+(2-2^{51})$