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 rest**2**. 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**.