Question

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

Answer 1

Hint: First we’ll create an equation for the division of ${x^{100}}$ by ${x^2} - 3x + 2$ using the remainder theorem which is $N = pq + r$ where N is divided by a number p the quotient and the remainder that results are q and r respectively, once we get the equation we’ll we have to find eliminate other factors in-order-to that we’ll substitute the value of x as one of the roots of dividend to get the equation in a variable as remainder.

Complete step-by-step answer:
Given data: ${x^{100}}$ and ${x^2} - 3x + 2.$
According to the remainder theorem, when any number N is divided by a number p the quotient and the remainder that results are q and r respectively
Then, $N = pq + r$
Now, according to the given condition, we can say that N is ${x^{100}}$and p is ${x^2} - 3x + 2.$
Now let $r = lx + m$ because while dividing a higher degree polynomial to a lower degree polynomial the maximum degree of the remainder is 1.
Now, factorizing for p
i.e. ${x^2} - 3x + 2$
Now dividing the coefficient of the first-degree term in such a way that they are the factors of the coefficient of other two terms,
$ \Rightarrow {x^2} - \left( {2 + 1} \right)x + 2$
Now, on simplifying the brackets, we get,
$ \Rightarrow {x^2} - 2x - 1x + 2$
Now, taking common from the first two and last two terms,
$ \Rightarrow x(x - 2) - 1(x - 2)$
Taking (x-2) common, we get,
$ \Rightarrow (x - 2)(x - 1)$
Now, substituting the value of N,p and r in $N = pq + r$
$ \Rightarrow {x^{100}} = (x - 2)(x - 1)q + lx + m$ ………(1)
Substituting x=1, we get,
$ \Rightarrow 1 = l + m..........(i)$
Substituting x=2 in equation (1), we get,
$ \Rightarrow {2^{100}} = 2l + m..........(ii)$
Subtracting equation(i) from equation(ii), we get,
$ \Rightarrow {2^{100}} - 1 = l$
Substituting the value of l in equation(i), we get,
$ \Rightarrow 1 = {2^{100}} - 1 + m$
$ \Rightarrow m = 2 - {2^{100}}$
Therefore, the remainder is lx+m
i.e. $\left( {{2^{100}} - 1} \right)x + 2 - {2^{100}}$.

Note: We can verify our above answer by taking an example or we can say that substituting the value of x.
Let $x = 1$
i.e. $N = 1$
$p = \left( {x - 1} \right)\left( {x - 2} \right)$
$\therefore p = 0$
Using remainder theorem
i.e. $N = pq + r$
$ \Rightarrow 1 = 0q + r$
$\therefore r = 1$
And substituting x=1 in the remainder that we’ve found as the result
i.e. $\left( {{2^{100}} - 1} \right)x + 2 - {2^{100}}$
$ \Rightarrow {2^{100}} - 1 + 2 - {2^{100}}$
$ \Rightarrow 1$ and hence answer is similar to the remainder resulting according to the remainder theorem.