Factor Representation Of A Polynomial

Theorem: If $\alpha_1, \alpha_2, \cdots , \alpha_m$ are the roots of a polynomial equation $f(x)=0$ and $f(x)$ has a degree $n$ $(n>m)$, then $f(x)$ can be written as $(x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_m) g(x)$, such that $deg\big(g(x)\big) = n-m $

Prerequisites:
Factor Theorem (proof)

Proof:
If $\alpha_1$ is a root of the polynomial $f(x)$, then by Factor Theorem, $f(x)$ can be written as:

$\qquad\quad f(x) = q_1(x)(x-\alpha_1)\qquad\qquad\qquad\qquad\qquad\cdots (1)$

Now, since $\alpha_2$ is also a root of $f(x)$, hence $f(\alpha_2) = 0$

$\therefore\quad\;\;\; f(\alpha_2) = q_1(\alpha_2)(\alpha_2-\alpha_1)=0$

Since, $\alpha_2\neq\alpha_1$, therefore $q_1(\alpha_2) = 0$.

Now using Factor Theorem on $q_1(x)$:

$\qquad\quad q_1(x) = q_2(x)(x-\alpha_2)$

Substituting this in equation $(1)$,

$\qquad\quad f(x) = q_2(x)(x-\alpha_2)(x-\alpha_1)$

Continuing in this way, we obtain the desired result:

$\qquad\quad f(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_m) g(x)$

Degree of g(x):

$\qquad\quad deg\big(LHS\big) = n$
$\qquad\quad\!\begin{align}deg\big(RHS\big) &= deg\big((x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_m)\big) + deg\big(g

(x)\big)\\
&= m + deg\big(g(x)\big)\\
&= deg\big(LHS\big) = n\end{align}\\
\Rightarrow\quad\;\; deg\big(g(x)\big) = n-m$

Hence the result.

Corollary:
(i) Maximum number of real roots of a polynomial of degree $n$ can be $n$.
$\quad$This follows since, $deg\big(g(x)\big) = n-m > 0$, implying $m < n$.

(ii) Total number of all real and complex roots of a polynomial of degree $n$ are $n$.


Recommended:
Sum and product of the roots
Condition for roots of quadratic equation
Quadratic Formula