Proof strategy polynomial division

Question

I'm given the following polynomial

$$ p(x) = x^4 -2x^3 +6x^2-10x+5 \in \mathbb{Q}[x] $$

I need to prove that $(x-1)^2|p(x)$.

Which are the possible ways to prove that?

One way is to do the polynomial long division and check if the remainder is 0.

Are there other ways for example by observing that 1 is a root?

Answer 1

Observing that $1$ is a root proves only that $x-1\mid p(x)$, not that $(x-1)^2\mid p(x)$. However, if you additionally verify that $1$ is a root of the derivative $p'(x)$, that does prove that $(x-1)^2\mid p(x)$. In general, $(x-a)^n$ divides a polynomial $p(x)$ iff $p^{(k)}(a)=0$ for $k=0,\dots,n-1$. (This follows, for instance, from considering the Taylor expansion of $p$ at $a$. Or, it is obvious when $a=0$ since both conditions are equivalent to $p$ having no terms of degree less than $n$, and then you can apply this to $p(x+a)$ to handle general $a$.)

Answer 2

Well, polynomial division is a nice exercise. Our Eileen knows it very well in high school now.

Another strategy is to substitute 1 into $p(x)$ and calculate $p(1)$. If $p(1)=0$, then you already have a factor $x-1$. Next, calculate the derivative $p'(x)$. If 1 is a zero, you have a double root 1. Then iterate.

Answer 3

If $(x-1)^2$ divides it then, $x=1$ is a root, so if we sub $x=1$ into the equation we want $p(x)=0$. Which is correct if you try it yourself. Or you could notice that $p(x)=(x-1)^2(x^2+5)$, so clearly $(x-1)^2$ divides $p(x)$!

Answer 4

You need to show that $p(1)=p'(1)=0$

Note that for $$ p(x) = x^4 -2x^3 +6x^2-10x+5 $$

$$p(1) = 1-2+6-10+5 =0$$

Also $$p'(x) = 4x^3 -6x^2 +12x-10$$ and $$p'(1) = 4-6+12-10 =0$$

Answer 5

Another possible way is to shift the polynomial setting $\boxed{y=x-1\Leftrightarrow x=y+1}$ and to use

• $(x-1)^2|p(x) \Leftrightarrow y^2|p(y+1)$ and
• $(1+y)^n = 1+ny+y^2q_n(y)$ for $n\geq 2$ where $q_n(y)$ is polynomial resulting from the binomial expansion.

Hence \begin{eqnarray*} p(y+1) & = & (1+y)^4-2(1+y)^3 +6(1+y)^2-10(1+y)+5 \\ & = & \color{blue}{1+4y} +y^2 q_4(y) \color{blue}{-2 - 6y}+y^2q_3(y)\color{blue}{+ 6+12y} + y^2\color{blue}{-10-10y+5} \\ & = & \color{blue}{0} +y^2Q(y) \mbox{ with a polynomial } Q(y)\\ \end{eqnarray*} It follows $\boxed{y^2|p(y+1)}$. Done.