# Proof of Different Polynomial Decompositions into Linear Factors

From G. Polya "Mathematics and Plausible Reasoning" p. 18.

How do you prove that provided the roots of a polynomial are different from zero,

$$a_0 + a_1x+a_2x^2 + ... + a_nx^n$$

$$\,= a_0\left(1-\frac{x}{\alpha_1}\right)\left(1-\frac{x}{\alpha_2}\right)...\left(1-\frac{x}{\alpha_n}\right)$$ with $\alpha_1, \alpha_2,...\alpha_n$ corresponding to the polynomial roots.

Setting $x=0$ results in:

$\qquad a_0+a_1x+\ldots+a_nx^n\,=\,(-1)^n\,a_n\alpha_1\cdot\ldots\cdot\alpha_n\left(1-\frac{x}{\alpha_1}\right)\cdot\ldots\cdot\left(1-\frac{x}{\alpha_n}\right)$

simplifying to

$$a_0=(-1)^n\,a_n\alpha_1\cdot\ldots\cdot\alpha_n$$ Therefore:

$\qquad a_0+a_1x+\ldots+a_nx^n \,=\,a_0\left(1-\frac{x}{\alpha_1}\right)\cdot\ldots\cdot\left(1-\frac{x}{\alpha_n}\right)$

• Great. Thank you. Before I "accept" / click on the check mark, can I tweak a bit your response for completeness, you approve it, and we close it? – Antoni Parellada May 30 '15 at 22:13
• @AntoniParellada, make the edit you feel required. – enzotib May 30 '15 at 22:14
• It truly wasn't necessary... Feel free to erase the changes. – Antoni Parellada May 30 '15 at 22:27
• @AntoniParellada: it is ok. – enzotib May 30 '15 at 22:27
• I made some changes to splice this post with a proof of Viete's theorem I am working on. It's the same without the $k$ step. – Antoni Parellada May 31 '15 at 0:37

It's not clear what you actually want shown. That the polynomial has $n$ roots is a consequence of the fundamental theorem of algebra. That the factors $(1-x/\alpha_i)$ have the form they do is actually the factor theorem in a different form. Then the left-hand side divided by all the $(1-x/\alpha_i)$s is a polynomial of degree $0$, i.e. constant. You can therefore find the constant by setting $x=0$, so all that is left is $a_0/1=a_0$.

• Yes. The issue is with the way the factors are expressed, as opposed to $(x -\alpha_i)$. Any possibility you can unfold the reasoning a bit more? – Antoni Parellada May 30 '15 at 21:52
• Okay, one possibility is to put $y=1/x$, and then you have a rational function $\frac{1}{y^n}(a_0 y^n + a_1 y^{n-1}+ \dotsb + a_n)$, on which the normal factor theorem then works after multiplying by $y^n$, but the roots are then $y=1/\alpha_i$, obviously, so you have factors of the form $(y-1/-\alpha_i)$ You can then re-substitute and push the $x^n$ into the factors $(1/x-1/a_i)$ to get the form specified. – Chappers May 30 '15 at 22:00

We assume that we are working in the complex numbers, or the reals, or the rationals.

Let $P(x)$ be the first polynomial, and let $Q(x)$ be the second. Each has degree $n$, and they have the same roots, counting multiplicity.

Let $D(x)=P(x)-Q(x)$. Then $D(x)$ has degree $\le n$. Note that $D(x)$ has at least $n+1$ roots, since the constant term of $D(x)$ is $0$.

Any polynomial of degree $\le n$ which has $\ge n+1$ roots, counting multiplicity, is the zero polynomial. The result follows.

As an appendix to the initial question, and as a matter of conceptual proximity in the derivation of Basel problem solution the proof offered as a response can be applied to the following situation:

If the polynomial is as follows:

$$b_0 - b_1\, x^2 +b_2\,x^4-\cdots+(-1)^n\,b_n\,x^{2n},$$

there will be $2n$ roots coming in pairs $β_i$ and $−\,β_i$.

As before the factorial decomposition can be expressed as the product of linear factors:

$$b_n\,\left(x-\beta_1\right)\,\left(x+\beta_1\right)\,\left(x-\beta_2\right)\left(x+\beta_2\right)\, \cdots\,\,\left(x-\beta_n\right)\left(x+\beta_1\right)$$

And multiplying numerator and denominator by $\frac{\beta_i^2}{\beta_i^2}$,

$$b_n\,\frac{\beta_1^2}{\beta_1^2}\left(x-\beta_1\right)\,\left(x+\beta_1\right)\,\,\frac{\beta_2^2}{\beta_2^2}\left(x-\beta_2\right)\left(x+\beta_2\right)\, \cdots\,\,\frac{\beta_n^2}{\beta_n^2}\left(x-\beta_n\right)\left(x+\beta_n\right)$$

which simplifies to:

$$b_n\,\beta_1^2\,\beta_2^2\,\cdots\,\beta_n^2\left(\frac{x}{\beta_1}-1\right)\,\left(\frac{x}{\beta_1}+1\right)\left(\frac{x}{\beta_2}-1\right)\, \left(\frac{x}{\beta_2}+1\right)\cdots\,\left(\frac{x}{\beta_n}-1\right)\left(\frac{x}{\beta_n}+1\right)$$

and changing the order within the parentheses:

$(-1)^n\,b_n\,\beta_1^2\,\beta_2^2\,\cdots\,\beta_n^2 \left(1-\frac{x}{\beta_1} \right)\,\left(1+\frac{x}{\beta_1}\right)\left(1-\frac{x}{\beta_2}\right)\, \left(1+\frac{x}{\beta_2}\right)\cdots\,\left(1-\frac{x}{\beta_n}\right)\left(1+\frac{x}{\beta_n}\right)$ or $\\$

$(-1)^n\,b_n\,\beta_1^2\,\beta_2^2\,\cdots\,\beta_n^2 \left(1-\frac{x^2}{\beta_1^2} \right)\,\left(1-\frac{x^2}{\beta_2^2}\right)\, \cdots\,\left(1-\frac{x^2}{\beta_n^2}\right).$

As before when $x=0$, we'll prove that $b_0=(-1)^n\,b_n\,\beta_1^2\,\beta_2^2\,\cdots\,\beta_n^2.$ Hence,

$$b_0 - b_1\, x^2 +b_2\,x^4-\cdots+(-1)^n\,b_n\,x^{2n}=b_0\left(1-\frac{x^2}{\beta_1^2}\right)\,\left(1-\frac{x^2}{\beta_2^2}\right)\, \cdots\,\left(1-\frac{x^2}{\beta_n^2}\right).$$