If a polynomial $g(x)$ is equal to zero at $a$ ( $g(a) = 0$ ) then we can write $g(x) = (x - a)^k\cdot g_1(x)$.

Question

It's written in my book that $g(x)$ is polynomial and if it's equal to zero at $a$, that is $ g(a) = 0 $ then we can write $g(x) = (x - a)^k\cdot g_1(x)$. Well, this proposition popped in my book without any description or derivation. Can anybody give me an intuition of what it means and derive it? Furthermore, I read this in the portion "Limits of rational functions". What this expression could mean in that context?

Answer 1

This follows from the Euclidean division of polynomials, which implies $$ g(x) = (x-a)q(x)+g(a) $$ Thus, $a$ is a root of $g$ iff $x-a$ divides $g$ and so $g(x)=(x-a)q(x)$.

Now repeat the argument with $q$. If $a$ is not a root of $q$, then we're done. If $a$ is a root of $q$, then $q(x)=(x-a)q_2(x)$ and so $g(x)=(x-a)^2q_2(x)$.

This process cannot go on forever because the degree of the quotient is decreasing. We end up with $g(x)=(x-a)^k q_k(x)$ with $q_k(a)\ne 0$.

Answer 2

This can be proven simply using the Euclidean division Algorithm.

For example,

$g(x)=x^3+x^2+3x+3$, and $g(-1)=0$, so you can divide $x^3+x^2+3x+3$ with $x+1$

Using Euclidean division you will get :$\frac{x^3+x^2+3x+3}{x+1}=x^2+3$

So you can write $g(x)=(x+1)(x^2+3)$

Note: The $k$ you mentioned is the multiplicity of the zero/root. In this case the multiplicity is $1$ so $k=1$

Answer 3

Well, let $$g(x)=g_0+g_1x+\dots+g_nx^n$$ be a polynomial of degree $n$ - so $g_n\neq0$.

Write down the euclidean division of $g(x)$ with the polynomial $(x-a)$: $$g(x)=(x-a)p(x)+\pi(x)$$ where $\deg\pi(x)<\deg(x-a)=1$ or $\pi(x)=0$. So $\pi(x)=c$, where $c$ is a constant. Note also that $$g(a)=0\Leftrightarrow\pi(a)=0\Leftrightarrow c=0$$ So, we have that: $$g(x)=(x-a)p(x)$$ So, the wanted result has been proved - with $k=1$. Now, it is obvious that, if $p(a)=0$, then there exists another polynomial $p_1(x)$ such that: $$p(x)=(x-a)p_1(x)$$ So $$g(x)=(x-a)^2p_1(x)$$ where $\deg g(x)=n>\deg p(x)=n-1>\deg p_1(x)=n-2$, we can continue this way until we find a polynomial $p_k(x)$ that is not divisible by $(x-a)$, where $\deg p_k(x)=n-k-1$ and $k\in\{1,2,\dots,n-1\}$. Then we have that: $$g(x)=(x-a)^{k+1}p_k(x)$$.

Answer 4

If you want to go back to basics:

Define $f(x)=g(x+a)$. Then it's also true that $g(x)=f(x-a)$.

$f$ is a polynomial of the same degree as $g$ and $f(0)=g(0+a)=0$. That means its constant coefficient is $0$:

$$f(x)=a_nx^n+\cdots+a_2x^2+a_1x$$

Then we define

$$f_1(x)=a_nx^{n-1}+\cdots+a_2x+a_1$$

and we can see that $f(x)=xf_1(x)$. What follows is:

$$g(x)=f(x-a)=(x-a)f_1(x-a)$$

Calling $g_1(x)=f_1(x-a)$, we see that $g(x)=(x-a)g_1(x)$. Note that $g_1$ is one degree less than $g$.

If $g_1(a)$ is still $0$, we know already that $g_1(x)=(x-a)$ times another polynomial, let's call it $g_2(x)$, so $g(x)=(x-a)^2g_2(x)$. Repeating until we end up with non-zero $g_k$, we get

$$g(x)=(x-a)^kg_k(x)$$

Example: $g(x)=x^3-8$, $g(2)=0$.

Then we have

$$f(x)=g(x+2)=(x+2)^3-8=x^3+3\cdot2x^2+\cdot4x+8-8=x^3+6x^2+12x$$ $$f_1(x)=x^2+6x+12$$

You already knew the constant coefficient of $f$ would be zero because $f(0)=g(0+2)=0$. Then $f(x)=xf_1(x)$. Next we do

$$g_1(x)=f_1(x-2)=(x-2)^2+6(x-2)+12=\cdots=x^2+2x+4$$

The way we arrived at $g_1$, it should be no surprise (but try if you wish) that

$$g(x)=(x-2)g_1(x)=(x-2)(x^2+2x+4)$$

Because $g_1(2)\neq0$, we came to an end. In this example $k=1$.