On finding the zeros of a polynomial

Question

What is the zero (real) of the polynomial $$x^{k+1}-2x^{k}+1=0$$ If there is such, how can I find it or what method can I use?

Answer 1

Well, one thing you can observe is that $x=1$ is a zero, as is always the case when the coefficients add up to $0.$ At that point, you could use polynomial long division or synthetic division to factor.

Alternately, note that the factorization is easy for $k=0$ and $k=1,$ and for $k\ge 2$ we have $$\begin{align}x^{k+1}-2x^k+1 &= x^{k+1}-x^k-x^k+1\\ &= x^k(x-1)-(x^k-1)\\ &= x^k(x-1)-(x-1)\sum_{j=0}^{k-1}x^j\\ &= (x-1)\left(x^k-\sum_{j=0}^{k-1}x^j\right).\end{align}$$

Sadly, finding the zero(es) (aside from $x=1$) of this polynomial will be impossible to determine in general for sufficiently large $k.$ It's easy for $k=2,$ but after that, it quickly goes from tedious to impossible to determine analytically (the best we're likely to be able to do for most $k$ is find numerical approximations).

Answer 2

Let $f(x)=x^{k+1}-2x^k+1$. For $k=1$ the polynomial is just $(x-1)^2$ and has $1$ as double root. For $k=2$ we find that $f(x)=(x-1)(x^2-x-1)$ has roots $1$ and $\frac{1\pm\sqrt 5}{2}$. So now assume $k\ge3$. From $$\tag1f(x)=x^{k+1}-2x^k+1 =(x-1)(x^k-x^{k-1}-x^{k-2}-\ldots-1)$$ we see that - again - one simple root is at $x=1$. For $x\ge2$ we have $$\tag2 f(x)=x^{k-1}((x-1)^2-1)+1\ge 1$$ and for $x\le-1$ we have $|f(x)|=|x^{k-1}((x-1)^2-1)+1|\ge |x^{k-1}((x-1)^2-1)|-1\ge 2$, hence all real roots are in the interval$(-1,2)$. For $1-\sqrt 2\le x<1$, we see from $(2)$ that $$|f(x)-1|\le|(x-1)^2-1|<1,$$ hence there are no roots in $[1-\sqrt 2,1)$.

The second factor in $(1)$ is $=1-k<0$ at $x=1$, hence $f(1+h)<0$ for small positive $h$, hence there is (at least) another root in $(1,2)$, which is best found numerically. If $k$ is even, $f(-1)=-2<0<1=f(0)$ shows that a third real root must be in $(-1,0)$, also best found by numerical methods.

We have $$ f'(x) = (k+1)x^k-2kx^{k-1} = ((k+1)x-2k)x^{k-1}$$ which has a $(k-1)$fold root at $0$ and a simple root at $\frac{2k}{k+1}=2-\frac2{k+1}$. Since between any two roots of $f$ there must be at least one root of $f'$, we conclude that

• There is only one root of $f$ in $(1,2)$ and it must be in $(2-\frac2{k+2},2)$
• There is at most one root of $f$ in $(-\infty,0)$.

Hence for $k$ even there are exactly three real roots, one in $(-1,1-\sqrt 2)$, one at $1$ and one in $(2-\frac2{k+2},2)$. And for $k$ odd there are exactly two real roots, one at $1$ and one in $(2-\frac2{k+2},2)$.