Let's calculate the rest of the division of :
R (x) = 3
The root of the divider is .
That is when B (x) is a polynomial of degree 1, the rest is equal to the numerical value of P (x) When x assumes the root value of B (x).
To demonstrate this fact, let's effect:
Note that the degree of rest is 0because it is smaller than the divider degree, which is 1. So the rest is a constant r.
Performing we have:
Thus, we can state the following theorem:
The rest of division of a polynomial P (x) by the binomial ax + b equals the numerical value of this polynomial for , that is, .
Calculate the remainder of the division of P (x) = x² + 5x - 1 per B (x) = x + 1:
We found the root of the divisor:
x + 1 = 0 x = - 1
From the rest theorem, we know that the rest is equal to P (-1):
P (-1) = (-1) ² + 5. (-1) -1 P (- 1) = - 5 = r
So the rest of the division of x² + 5x - 1 per x + 1 é - 5.
note that P (x) is divisible by ax + b When r = 0, ie when . Hence comes the statement of the following theorem:
A polynomial P (x) is divisible by the binomial 1 if and only if .
The most important case of division of a polynomial P (x) is the one where the divisor is of the form (x - ).
note that is the root of the divisor. So the rest of the division of P (x) by (x -) é:
r = P ()
P (x) is divisible by (x - ) When r = 0, ie when P() = 0.
Determine the value of Pso that the polynomial be divisible by x - 2:
For what P (x) be divisible by x - 2 we should have P (2) = 0, because 2 is the root of the divisor:
So for that be divisible by x - 2 we should have p = 19.Next: Division of a Polynomial by (x-a) (x-b)