More fun in 2015, Part 4

Let ${(1+x)}^{2015}=h(x)*(x^{2}+x+1)+ ax+b$

Where h(x) is the polynomial quotient and ax+b is the polynomial remainder( Dividing a polynomial by a quadratic yields a linear remainder)

Now $x^{2}+x+1= (x+\omega)(x+{\omega}^{2})$

Where $\omega, {\omega}^{2}$ are non real cube roots of unity.

In the above expression put $x=\omega$ to get

$-\omega= a\omega+b$

Comparing real and imaginary parts we have

$a=-1,b=0$

Remaimder = -x

Solution as Ronak Agarwal but different in presentation.

Using Remainder Theorem, we have:

$(1+x)^{2015} = Q(x)(x^2+x+1) + R(x)$ , where $R(x)$ is the remainder.

We note that the roots of $x^2+x+1 = 0$ are the complex cubic roots of $1$ that is $\omega^2 +\omega + 1 = 0\Longrightarrow \omega^2 = -(1+\omega)$ and $\omega^3 = 1$ .

Therefore,

$\begin{aligned} (1+\omega)^{2015} & = Q(\omega)(\omega^2 +\omega + 1) + R(\omega) \\ (-\omega^2)^{2015} & = 0 + R(\omega) \\ -\omega^{4030} & = R(\omega) \\ -\omega^{3\times 1343 + 1} & = R(\omega) \\ -\omega & = R(\omega) \end{aligned}$

$\Rightarrow R(x) = \boxed{-x}$

We know that the remainder is a lineal function.

Let $R(x) = ax + b$

We only have to know two points of the line.

Then $R(0) = a\cdot 0 + b = b = 0$ (calculating the remainder when $x=0$ ).

Therefore $R(1) = a\cdot 1 + b = a = -1$ (calculating the remainder when $x=1$ ).

Then $\Rightarrow R(x) = a\cdot x + b = -1\cdot x + 0 = -x$