Polynomial Divisibility Problem: Proof & Corollary

[kaliprasad]

problem
For any polynomial P(x) show that P(a) - P(b) is divisible by a-b
Proof:
Let $p (x) = t_nx^n + t_{n-1} x^{n-1} + \cdots + t_0$
Then
$p (a) = t_na^n + t_{n-1} a^{n-1} + \cdots + t_0$
$p (b) = t_nb^n + t_{n-1} b^{n-1} + \cdots + t_0$
So $p (a) – p(b) = t_n(a^n- b^n) +t_{n-1} (a^{n-1}-b^{n-1}) + \cdots + t_1(a-b)$
As each of the $a^k-b^k $ is divisible by a- b so p(a) – p(b) is dibvisible by a-b.
As a corollary
If p(x) has integer coefficients and P(0) and P(1) are odd it does not have any integer root.
This is so because P(even) – p(0) is even and P(odd) – p(1) is even so neither can be zero.

 

[zzephod]

problem
For any polynomial P(x) show that P(a) - P(b) is divisible by a-b
Proof:
Let $p (x) = t_nx^n + t_{n-1} x^{n-1} + \cdots + t_0$
Then
$p (a) = t_na^n + t_{n-1} a^{n-1} + \cdots + t_0$
$p (b) = t_nb^n + t_{n-1} b^{n-1} + \cdots + t_0$
So $p (a) – p(b) = t_n(a^n- b^n) +t_{n-1} (a^{n-1}-b^{n-1}) + \cdots + t_1(a-b)$
As each of the $a^k-b^k $ is divisible by a- b so p(a) – p(b) is dibvisible by a-b.
As a corollary
If p(x) has integer coefficients and P(0) and P(1) are odd it does not have any integer root.
This is so because P(even) – p(0) is even and P(odd) – p(1) is even so neither can be zero.

Let $$Q_b(x)=P(x)-P(b)$$ this has a root at $$x=b$$ so there exists a polynomial $$R_b(x)$$ such that:

$$Q_b(x)=(x-b)R_b(x)$$,

and in particular:

$$Q_b(a)=P(a)-P(b)=(a-b)R_b(a)$$

.