Remainder of polynomial division

use remainder theorem. Substitute 1 to the equation

Use the polynomial remainder theorem:

${ x }^{ 2015 }+1=(x-1)(D(x))+R(x)$ , basically the expression is equal to $x-1$ multiplied by some other number and add the remainder.

Substitute $1$ for $x$ so that $(x-1)(D(x))$ would equal to $0$ . Doing so will leave us with the remainder:

${ 1 }^{ 2015 }+1=(1-1)(D(1))+R(1) \longrightarrow{1+1=R(1)} \longrightarrow{R(1)=2}$

Therefore the remainder is $\boxed{2}$

Let $P(x) = x^{2015} + 1$ and $Q(x) = x - 1$ . We want to find the remainder of $\frac{P(x)}{Q(x)}$ .

Now, we can write $P(x)$ as the sum of two polynomials $R(x)$ and $S(x)$ , such that: $R(x) = x^{2015} - 1$ and $S(x) = 2$ . Notice that, since $R(x)$ has $1$ as a root, it can be divided by $Q(x)$ ; thus, $P(x) \equiv S(x) \pmod{x - 1}$ , and so 2 is the remainder sought.

Let $q(x),r(x) \in \mathbb C[X]$ be the quotient and remainder of this division. respectively Then, we may write : $x^{2015}+1 = q(x)(x-1)+r(x) \quad (\dagger)$ The trick is NOT to try figuring out what $q(x)$ is. To be honest, even as I finished setting up this little exercise, I had (and still have) no idea what the quotient looks like, nor am I sufficiently bored to calculate it!

All that we need to notice is that $\deg(r) < 1$ , which is to say that $r(x)$ is a constant! So, we might as well set : $r(x) \equiv c$ .

Since the value is the same $\forall x$ , we can evaluate $(\dagger)$ for $x \equiv 1$ .

We easily get that : $c=\boxed{2}$

however 1, 2, 3 shall not satisfy the value of x

x - 1 definitely divides x to the 2015 minus 1. Therefore x 2015 + 1 has a remainder of 2.

if x-1 divides p(x)= x^2015 then put x-1=0 .it implies x=1. put x=1 in p(x) .then answer is 1.

just put one for x and add a negative sign, then add