Q:

Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$

Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$.
Let $P(x)=x^{2015}-x^{2014}=Q(x)(x-1)^3+ax^2+bx+c.$ If we put $x=1$ in $P(x)$ and $P'(x)$, we get $a+b+c=0$ and $2a+b=1$. Then: $c=a-1$. The second derivative won't help in finding $b$, so, what should I do? Thank you

A:

Rewrite expression as
$$
\frac{x^{2015} - x^{2014}}{(x-1)^3} = \frac{x^{2014}}{(x-1)^2}
$$
so we have
$$
x^{2014} = Q(x) (x-1)^2 + ax + b.
$$
Now you need to find coefficients $a$ and $b$. Your argument with derivatives should work:
$$
1^{2014} = 1 = a + b
$$
$$
2014\cdot 1^{2013} = 2014 = a.
$$
Hence, $b = -2013$ and the final answer is
$$
x^{2015} - x^{2014} = x^{2014} (x-1) = Q(x) (x-1)^3 + (x-1)\cdot(2014x - 2013)
$$

A:

$\dfrac{x^{2014}(x-1)}{(x-1)^3}= \dfrac{x^{2014}}{(x-1)^2}$;
$x^{2014}= Q(x)(x-1)^2+ ax+b$;
Binomial expansion:
$x^{2014}=(1+(x-1))^{2014}=$ 
$\sum_{k=0}^{n}\binom{2014}{k}1^{n-k}(x-1)^k$
Remainder $ax+b$:
$\binom{2014}{0}1+\binom{2014}{1}(x-1)=$
$1+2014x-2014=2014x-2013$.
Originally: 
$x^{2014}(x-1)$ is divided by $(x-1)^3$:
Hence 
$x^{2014}(x-1)=Q(x)(x-1)^3+(x-1)(2014x-2013),$
with quadratic remainder 
$(x-1)(2014x-2013)$.

