# 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

• Admittedly I haven't done out the full calculation, but doesn't the second derivative give you $a$? Then you could substitute to find $b$ and $c$. – Cade Reinberger Aug 22 at 21:34
• No, because $P''(x)=Q''(x)(x-1)^3+Q'(x)3(x-1)^2+Q'(x)3(x-1)+Q(x)3+2a$, and we don't know any value of $Q(x)$. – user672596 Aug 22 at 21:36
• This may help $\frac{x^{2015} - x^{2014}}{(x-1)^3} = \frac{x^{2014}}{(x-1)^2}$ – Virtuoz Aug 22 at 21:36
• @karim-ashli you have only two coefficients then, so you argument with derivatives will work – Virtuoz Aug 22 at 21:39
• If $x^{2014}=S(x)(x-1)^2+qx+r$ then $x^{2015}-x^{2014}=S(x)(x-1)^3+(qx+r)(x-1)$ – Mark Bennet Aug 22 at 21:42

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 $$x^{2015} - x^{2014} = x^{2014} (x-1) = Q(x) (x-1)^3 + (x-1)\cdot(2014x - 2013)$$

$$\dfrac{x^{2014}(x-1)}{(x-1)^3}= \dfrac{x^{2014}}{(x-1)^2}$$;

$$x^{2014}= Q(x)(x-1)^2+ ax+b$$;

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

$$x^{2014}(x-1)=Q(x)(x-1)^3+(x-1)(2014x-2013),$$

$$(x-1)(2014x-2013)$$.

Using $$\ zf\,\bmod zg\ = z\, (f \bmod g)\, =\,$$ mod Distributive Law to factor out $$\, z = x\!-\!1$$

\begin{align} z(z\!+\!1)^{\large n} \bmod z^{\large 3} &=\, z\,(\color{#c00}{(1+ z)^{\large n}}\bmod \color{#c00}{z^{\large 2}})\\[.4em] &=\,z\,(\color{#c00}{1+nz})\ \ \text{ by } \color{#c00}{\text{Binomial or Taylor }}\text{Theorem}\\[.4em] &=\ n\!-\!1 + (1\!-\!2n)x + n x^{\large 2}\ \ \ \ \ \text{[OP is }\, n = 2104] \end{align}

Another way:

Set $$x-1=y$$

$$x^r=(1+y)^r\equiv1+\binom r1y+\binom r2y^2\pmod{y^3}$$

$$x^m-x^n\equiv y(m-n)+y^2\left(\binom m2-\binom n2\right)\pmod{y^3}$$

Replace $$y$$ with $$x-1$$

• (+1) This is the way I did this. You can do it in your head this way. – robjohn Aug 23 at 9:09

I can use the division algorithm of a polynomial by another polynomial: in particular set $$A(x)=x^{2014}$$ and $$B(x)=x^2-2x+1$$. Proceeding with successive division, I obtain as quotient $$Q(x)=x^{2012}+2x^{2011}+3x^{2010}+\cdots + 2012x+2013$$ The general term of this polynomial is $$nx^{2013-n}$$. When $$n=2013$$, I obtain as polynomial rest: $$R(x)=-2012x+4026x-2013=2014x-2013$$. Before starting this process I have cancel out $$(x-1)$$ from the numerator and denominator of the fraction, so: $$Q(x)=(x^{2012}+2x^{2011}+3x^{2010}+\cdots + 2012x+2013)(x-1)$$ and $$R(x)=(2014x-2013)(x-1)$$

It is: $$\frac{x^{2015}-x^{2014}}{(x-1)^3}=\frac{(x-1+1)^{2015}-(x-1+1)^{2014}}{(x-1)^3}=\\ =\small{\frac{\left[S(x)+{2015\choose 2}(x-1)^2+{2015\choose 1}(x-1)+1\right]-\left[T(x)-{2014\choose 2}(x-1)^2-{2014\choose 1}(x-1)-1\right]}{(x-1)^3}}=\\ =\frac{S(x)-T(x)+2014(x-1)^2+(x-1)}{(x-1)^3}=Q(x)+\frac{(x-1)(2014x-2013)}{(x-1)^3}.$$ Note: It is more expanded and direct version of lab bhattacharjee's answer.