The polynomial coefficients divided by gaps
Coefficient of C in a look binomial X-C
The set polynomial has an appearance
if we divide it
Let's receive a polynomial
and rest

Consider the division procedure of a polynomial of the form

\[f(x)=a_0x^{n}+a_1x^{n-1}+a_2x^{n-2}+.....+a_{n-1}x+a_n\]

on a bin of the form $$x-c$$

the result of division is a function of the form

$$q(x)=b_0x^{n-1}+b_1x^{n-2}+b_2x^{n-3}+.....+b_{n-2}x+b_{n-1}$$

Such a result is obtained only by dividing the original polynomial by a bin without a remainder.

In the general case, it is said that the function $$f(x)$$ can be represented as $$f(x)=q(x)(x-c)+r$$

where r is the remainder of the division.

Function coefficients q (x) calculated by recurrence formulas

$$b_0=a_0\\b_1=cb_0+a_1\\...\\b_{n-1}=cb_{n-2}+a_{n-1}]$$

 $$r=cb_{n-1}+a_n$$

Horner's circuit is very convenient with its simple and lack of division function. This allows us to solve similar equations with increased accuracy, as well as to solve integer equations, without any machine (computer) errors.

By the way!

There is a new calculator that divides a polynomial into a polynomial with a remainder. It also works in a complex field, in addition, the dividing polynomial can actually be a polynomial (!), And not a binomial, as in this article.

In addition, the same scheme allows us to solve the problem of determining the value of a function for any value. "Fi!" - you say. "It's elementary, any calculator can do it."

 

yes of course, putting the necessary value instead of the unknown x, we get the desired result, but at what cost?

We will have to raise the values ​​to a power, which will undoubtedly introduce its own error into the calculations.

This is clearly manifested when working in the field of complex numbers, when dividing the polynomial into a complex bin.

It’s easier for us to use Bezout ’s theorem , which states: The remainder r from dividing the polynomial by f (x) on linear binomial $$x-c$$ equal to the value of the polynomial f (x) at x = c

The bot created on this site allows you to solve the problem by the Horner method, not only for real numbers, but also for complex ones. This extends the application of the bot and allows you to more fully explore the function.

If the dividing polynomial is not a monomial, then you should use a calculator that divides arbitrary polynomials into each other with the calculation of the remainder. Division of a polynomial into a polynomial. Division of complex polynomials Now let's look at examples.

$$f(x)=2x^3-3x+2$$ split with the remainder $$x+2$$

We write the coefficients 2 0 -3 2 and through the semicolon -2. I hope it is clear why we write -2, but not + 2?

We get the answer

The given polynomial has the form
$$f(x)=(2)*x^3+(-3)*x^1+(2)$$
if we divide it
$$x-(-2)$$
We get the polynomial
$$q(x)=(2)*x^2+(-4)*x^1+(5)$$
and the remainder
$$r=-8$$

 

The following example, the original polynomial is the same, but the value of C will be complex, for example 1 + i

We write the coefficients 2 0 -3 2 and through the semicolon 1 + i

We get

The given polynomial has the form
$$f(x)=(2)*x^3+(-3)*x^1+(2)$$
if we divide it
$$x-(1+i)$$
We get the polynomial
$$q(x)=(2)*x^2+(2+2i)*x^1+(-3+4i)$$
and the remainder
$$r=-5+i$$

 

Thus, we can write any values, including complex ones, in coefficients of both a divisible polynomial and a dividing binomial

Good luck!

 

 
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