# Division of Polynomials. Horner's Scheme. Bezout's Theorem

We can add, subtract, multiply and raise to the natural power the polynomials. We can divide the polynomial by polynomial sometimes.

If there is such polynomial `S(x)` , that `P(x)=Q(x)S(x)` , then we say, that the polynomial `Q(x)` divides polynomial `P(x)` and `P(x)` is called **dividend**, `Q(x)` is **divisor** and `S(x)` is **quotient**.

For example, the polynomial `P(x)=x^3-3x^2+5x-15` is divided by a polynomial `Q(x)=x^2+5`, because `x^3-3x^2+5x-15=(x^2+5)(x-3)`; in the quotient we obtain the polynomial `S(x)=x-3`.

If the polynomial `P(x)` isn′t divided by the polynomial `Q(x)`, then we consider the **division with remainder**. The possibility of such division follows from next property: for any two polynomials `P(x)` and `Q(x)` such, that degree `P(x)` isn′t less than degree `Q(x)` , exist only one pair of polynomials `S(x)` and `R(x)` such that the following holds

`color(blue)(P(x)=Q(x)S(x)+R(x))`, (1)

thus the degree of polynomial `R(x)` is less than the degree of polynomial `Q(x)` (the polynomial `R(x)` is called** remainder**).

For division of polynomials, that are reduced to the standart form, we use the rule of long division, that is similar to division of multi-digit numbers.

**Example 1**.We should divide `P(x)=3x^4+2x^3+70x^2+3x-4` by `Q(x)=x^2+5x+1`.

We should perform the long division:

So, `S(x)=3x^2-13x+132` is quotient, `R(x)=-644x-136` is remainder. The identity holds:

Let′s consider the process of division of n-th degree polynomial, that has the form of `P_n(x)=a_0x^n+a_1x^(n-1)+...+a_n`, by linear binomial expression `x-a` . Then we can perform the division with the special scheme, that is called <strong >Horner′s scheme.

In this case the identity (1) takes the form

`color(red)(a_0x^n+a_1x^(n-1)+...+a_n=(x-a)(b_0x^(n-1)+b_1x^(n-2)+...+b_(n-1))+r)`, (2)

where quotient has `(n-1)-th` degree and remainder has zero degree, i.e. is number. Since the polynomials on the left and right parts of identity (2) are same, then we open the brackets and obtain equalities, that express coincidence of coefficients for equal degrees of `x` :

`a_0=b_0` , i.e. `b_0=a_0` ,

`a_1=-ab_0+b_1`, i.e. `b_1=a_1+ab_0` ,

`a_2=-ab_0+b_1`, i.e. `b_2=a_2+ab_1` ,

`a_(n-1)=-ab_(n-2)+b_(n-1)` , i.e `b_(n-1)=a_(n-1)+ab_(n-2)` ,

`a_n=-ab_(n-1)+r`, i.e `r=a_n+ab_(n-1)` .

The calculation of coefficients of quotient and remainder place in the next table:

`a_0` | `a_1` | `a_2` | ... | `a_(n-1)` | `a_n` | |

a | `a_0=b_0` | `ab_0+a_1` | `ab_1+a_2` | ... | `ab_(n-2)+a_(n-1)` | `ab_(n-1)+a_n=r` |

We complete top row of table step-by-step; the bottom row, where are written coefficients of quotient and remainder, we complete gradually moving from left to right. In each cell of bottom row we write the sum of coefficients from the top row multiplyied by number `a`, that we obtain in the left cell of bottom row.

**Example 2**. Using the Horner′s scheme, we should divide `x^3+4x^2-3x+5` by `x-2`.

Let′s make the table:

1 | 4 | -3 | 5 | |

2 |

First step is:

1 | 4 | -3 | 5 | |

2 |

So, the quotient equals `x^2+6x+9` and the remainder equals `23`.

Let′s note, that the remainder of division of polynomial by binomial `x-a` we can find, don′t making division. It is correct the next property:

the remainder of division of polynomials `P(x)` by binomial `x-a` equals the value of polynomial for `x=a` (**Bezout theorem**).

So, in the example 2 we have `P(2)=2^3+4*2^2-3*2+5=23`.

From the **Bezout′s theorem** follow, that *polynomial `P(x)` devides by binomial `x-a` only then, when `a` is the root of this polynomial.*

**Example 3.** For what value of `lambda` does *the polynomial `P(x)=x^4+6x^2+lambdax+6` devide by binomial `x+2` ?*

*It is necessary, that the number -2 was the root of polynomial. We have `P(-2)=16+24-2lambda+6=46-2lambda`, i.e. the polynomial `P(x)` devides by `x+2` , when `&lambda=23` . *