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Division of a polynomial by (x - a) (x - b)


Let's suppose that a polynomial P (x) is divisible by (x - a) and by (x -b), being theB . Does P (x) is divisible by product (x - a) (x - b)?

The divider (x - a) (x - b) have degree 2, so the rest will be at most 1.

Then we can write:

How P (x) is divisible by (x - a), then P (a) = 0:

How P (x) is divisible by (x -b), then P (b) = 0:

We have the following system of equations:

Therefore, we can conclude that P (x) is divisible by (x - a) (x - b).

We thus have the following theorem:

If P (x) is divisible by (x - a) and by (x - b) with TheB, then P (x) is divisible by (x - a) (x - b).

Generalizing the theorem, if P (x) is divisible by with distinct then P (x) is divisible by .

Example 1

The polynomial is divisible by (x - 1) (x - 2)?

Resolution

If P (x) is divisible by (x - 1) and by (x - 2), then P (x) will be divisible by (x - 1) (x - 2).

P (x) is divisible by (x -1), this is, P (1) = 0?

- Yes.

P (x) is divisible by (x - 2), this is, P (2) = 0?

- Yes.

How P (x) is divisible by (x -1) and by (x - 2) So P (x) is divisible by (x - 1) (x - 2).

Example 2

A polynomial P (x), divided by x - 1, rest 4; divided by x +1, rest 2. What is the rest of the division of P (x) per (x - 1) (x + 1)?

Resolution

As the divider (x - 1) (x + 1) have degree 2, the degree of rest will be at most 1. We can write:

P (x) divided by x - 1 give rest 4. Like this, P (1) = 4:

P (x) divided by x + 1 give rest2. Like this, P (-1) = 2:

We have the following system of equations:

So the rest of the division of P (x) per (x - 1) (x + 1) é x + 3.

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