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Show that $(x-1)^2$ is a factor of $x^n -nx +n-1$

By factor theorem we know that $(x-a)$ is a factor of $f(x)$ if $f(a)=0$.

In this case, $f(x)=x^n -nx +n-1 \implies f(1)=0$

Hence we conclude that $(x-1)$ is a factor. From hereon, how can I say that $(x-1)^2$ is a factor?

Can we approach the problem without calculus approach? This problem was taken from a book of pre-calculus algebra.

Bill Dubuque
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Anurag Saha
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6 Answers6

4

Here is another elementary way using the binomial theorem

  • $(1+y)^n = \sum_{k=0}^n\binom{n}{k}y^k$

Set $\boxed{y=x-1}$ and note that

  • $(x-1)^2$ is a factor of $p(x) = x^n -nx +n-1$ if and only if $y^2$ is a factor of $p(y+1)$

Hence, \begin{eqnarray*} p(y+1) & = & (1+y)^n - (1+y)n + n-1\\ & = & 1+ny +\sum_{k=2}^n\binom{n}{k}y^k -n-ny+n-1 \\ & = & y^2\sum_{k=2}^n\binom{n}{k}y^{k-2} \end{eqnarray*} Done.

Bill Dubuque
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trancelocation
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  • That can be viewed as using the Binomial Theorem to do a Taylor expansion. Compared to the other ways its

    $$\large \begin{align} &g(0) = n\ \ {\rm for}\ \ g(y) = \dfrac{(y!+!1)^n-1}y\[.2em] \iff\ &f(1) = n\ \ {\rm for}\ \ f(x) = \dfrac{x^n-1}{x-1}\end{align}\qquad \qquad\qquad$$

     – Bill Dubuque May 17 '19 at 01:53
  • @BillDubuque Since the OP wanted a "pre-calculus" solution, it is not necessary to refer to Taylor, even if it can be considered to be Taylor. I was wondering that none of the answers presented the idea of shifting the polynomial which is an often used "algebra trick" when dealing with polynomials. So, I thought it was worth adding this here. – trancelocation May 17 '19 at 02:02
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Since calculus and double root test are not known, we can instead divide by $\,x\!-\!1\,$ twice as follows.

Note that $\,x\!-\!1$ divides $f(x)=(\color{#c00}{x^n-1})-\color{#0a0}n(x-1)\ $ by $\, f(1)= 0\,$ and the Factor Theorem.

$x\!-\!1$ divides $g(x) := \dfrac{f(x)}{x\!-\!1} = (\color{#c00}{x^{n-1}+\cdots+1})-\color{#0a0}n\ $ by $\ g(1) =n-n=0\,$ similarly.

So we infer $\,\ g(x) = \dfrac{f(x)}{x\!-\!1} = (x\!-\!1)\,h(x)\,$ for some polynomial $\,h(x).$

Hence we conclude $\ \ f(x)\, = (x\!-\!1)^2h(x)\ $ by scaling above by $\,x\!-\!1.$

Bill Dubuque
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3

Hint

$$x^n -nx +n-1=(x^n-1) -n(x -1)=(x-1)(x^{n-1}+x^{n-2}+...+x+1-n)$$

now, write

$$p(x)=x^{n-1}+x^{n-2}+...+x+1-n$$

and check that $p(1)=0$.

Arnaldo
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$$x^n-nx+n-1=(x^n-1)-n(x-1)$$ In precalculus algebra one should know the extremely useful fact that $$1+x+..+x^{n-1}=\frac{x^n-1}{x-1},$$ for $x$ not $1$. Then, multiplying by $1-x$ on both sides one gets $$(1+x+..+x^{n-1})(x-1)=x^n-1$$ and this holds for all $x$ (expand the lhs to see this). So $$x^n-nx+n-1=(x-1)(1+x+..+x^{n-1}-n)$$. Now write $n=1+...+1$ $n$ times. Then we can rewrite $$1+x+..+x^{n-1}-n=(1-1)+(x-1)+..+(x^{n-1}-1).$$ But each of the summands is divisible by $x-1$ because of the argument above with the geometric series! In fact, writing $$x^k-1=1+x+..+x^{k-1}$$ for each $k$ appearing in the summand, by dividing $1+x+..+x^{n-1}-n$ by $x-1$ one gets $$n-1+(n-2)x+(n-3)x^2+...+x^{n-2}$$ since every term $x^j$ will appear in $\frac{x^k-1}{x-1}$ with $k\leq n-1$ exactly when $j<k$and there are $n-1-j$ such terms. Putting everything together we get the polynomial in your question is $$(x-1)^2((n-1)+(n-2)x+...+x^{n-2}).$$ This is the most elementary solution I could think of.

M. Van
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$f(1)=f'(1)=0$ implies that $(x-1)^2$ divides $f$.

Tsemo Aristide
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By division theorem we have:

$$x^n -nx +n-1 = k(x)(x-1)^2+ax+b $$ So for $x=1$ we have $0 = 0+a+b$ so $a=-b$, so $$x^n -nx +n-1 = k(x)(x-1)^2+a(x-1) $$

$$(x-1)(x^{n-1}+...+x^2+x+1) -n(x-1)= k(x)(x-1)^2+a(x-1) $$

so, after dividing by $x-1$ we get $$(x^{n-1}+...+x^2+x+1) -n= k(x)(x-1)+a $$

Now puting $x=1$ again we have $$\underbrace{1+1+...+1}_n -n = 0+a\implies a=0$$

nonuser
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