Partial fraction decomposition Coefficients of a polynomial (numerator) through a gap Poles/roots of a polynomial (denominator) through a gap
 The set fractional and rational function It is displayed for the sum of simple multipliers with the following coefficients

In this article, the decomposition of a rational function into the sum of the simplest fractions will be considered and calculated.

The fractional rational function has the following form

$\frac{ax^m+bx^{m-1}+cx^{m-2}+.....+wx+r}{Ax^n+Bx^{n-1}+Cx^{n-2}+.....+Wx+R}$

For example, such

$\frac{x^2-3x+1}{x^4+x^3-8x^2-5}$

If the degree of unknown in the numerator is less than the degree of unknown in the denominator (as shown in the example), then such a function is called the correct fractional rational.

Each regular fractional rational function can be decomposed into the sum of the simplest fractions.

The simplest fraction has a general form $\frac{G}{(x+J)^d}$

For example  $\frac{2}{(x+3)}$ or $\frac{-8}{(x-4)^5}$

In almost every similar Internet resource, which discusses the decomposition of fractions, the method of uncertain coefficients is used as a solution method. We will not dwell on this method, since I do not want to produce another copy with slightly different text. We only recall that there it is necessary to solve a system of linear equations.

We will use a different technique, and the online calculator will also take this technique into service.

Source fraction

$\frac{x^2+x+2}{x^5-5x^4-21x^3+41x^2+68x-84}$

So, if we know all the roots of the denominator in our function, then we can convert

$\frac{x^2+x+2}{x^5-5x^4-21x^3+41x^2+68x-84}=\frac{x^2+x+2}{(x-1)(x+3)(x-7)(x+2)(x-2)}$

That is, we need to expand the function in the following form

$\frac{x^2+x+2}{(x-1)(x+3)(x-7)(x+2)(x-2)}=\frac{A}{(x-1)}+\frac{B}{(x+3)}+\frac{C}{(x-7)}+\frac{D}{(x+2)}+\frac{E}{(x-2)}$

and determine all unknown coefficients

We use the method (sometimes it is called the method of partial values), the practice of application is as follows:

Let x = 1 then substituting this value into the numerator we get the value 4, substituting the denominator without (x-1) we get (1 + 3) (1-7) (1 + 2) (1-2) = 72

Therefore, the coefficient $A=\frac{4}{72}=\frac{1}{18}$

Now let x = -3. Then the numerator is (-3) ^ 2 + (- 3) + 2 = 8, and the denominator is (-3-1) (- 3-7) (- 3 + 2) (- 3-2 ) = 200

Therefore, the coefficient $B=\frac{8}{200}=\frac{1}{25}$

Running through all x equal to 1, -3, 7, -2, 2,   we calculate the coefficients

$A=\frac{1}{18}$

$B=\frac{1}{25}$

$C=\frac{29}{1350}$

$D=-\frac{1}{27}$

$E=-\frac{2}{25}$

$\frac{x^2+x+2}{x^5-5x^4-21x^3+41x^2+68x-84}=\frac{-2}{25 (x - 2)} + \frac{1}{18 (x - 1)} - \frac{1}{27(x + 2)} + \frac{1}{25 (x + 3)} + \frac{29}{1350 (x - 7)}$

In my opinion, this solution is simpler. But this technique can be used when the denominator does not have multiple roots.

"How? Is such an easy way not applicable in the case of multiple roots?" - the reader exclaims distressedly.

Not everything is so bad really. Just in the case of multiple roots, for example, the   $\frac{x^2+x+2}{(x-1)^3(x+3)(x-7)^2}$ calculation is more complicated. I will explain the algorithm now.

But for this you must be able to calculate the derivative of the polynomial, I hope you can do it.

First, convert the denominator to a polynomial. In order not to multiply manually, we use  Create a polynomial at its roots

Let us introduce the roots of the denominator taking into account their multiplicity 1 1 1 -3 7 7

We get $x^6+(-14)*x^5+(43)*x^4+(92)*x^3+(-409)*x^2+(434)*x^1+(-147)$

We need to convert the initial fraction into this form

$\frac{x^2+x+2}{(x-1)^3(x+3)(x-7)^2}==\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}+\frac{D}{(x+3)}+\frac{E}{(x-7)}+\frac{F}{(x-7)^2}$

Based on the above example, we can immediately find out what the coefficients D, C and F are

$D=\frac{(-3)^2-3+2}{(-3-1)^3(-3-7)^2}=\frac{8}{-64*100}=-\frac{1}{800}$

$C=\frac{(1)^2+1+2}{(1+3)(1-7)^2}=\frac{4}{4*36}=\frac{1}{36}$

$F=\frac{7^2+7+2}{(7-1)^3(7+3)}=\frac{58}{216*10}=\frac{29}{1080}$

Let's try to find out the coefficient B

Take the first derivative of the numerator. She is equal  $2x+1$.

Substitute the unit there, divide by one factorial 1! = 1 and and  remember the  value = 3

Now the denominator. We find out the values ​​of the derivatives of the denominator (for x = 1) through the online service. Value of a derivative of a polynomial by Horner's method

By entering the coefficients of the polynomial $x^6+(-14)*x^5+(43)*x^4+(92)*x^3+(-409)*x^2+(434)*x^1+(-147)$

we get the table

Preset function
$f(x)=x^{6}+(-14)*x^{5}+(43)*x^{4}+(92)*x^{3}+(-409)*x^{2}+(434)x+(-147)$
 Derivative The value of the derivative at X = 1 0 0 1 0 2 0 3 864 4 -288 5 -960 6 720

The first and second derivatives are equal to zero, but this is logical, since root 1 has a multiplicity of 3.

The value of the third derivative is 864

We make the equation

3 (this is the numerator I asked to remember) = B * 864/3! (Three factorials) + C * (- 288) / 4!

While for the majority it is not at all obvious what and where, but don’t worry, knowledge will come.

C we know and get an equation with one unknown

3 = B * 864 / 6-288 / 36/24

$B=\frac{20}{864}=\frac{5}{216}$

Let's find out what is the coefficient A.

We have almost everything

We take the second derivative of the numerator. It is equal to a constant number = 2

Divide into two factorial 2! = 2 and and  remember the  value = 1

Now we make the equation

1 = A * 864/3! + B * (- 288) / 4! + C (-960) / 5!

B and C are known to us.

$1=144A-12B-8C=144A-12\frac{5}{216}-8\frac{1}{36}$

$A=\frac{1}{96}$

I.e

$\frac{x^2+x+2}{(x-1)^3(x+3)(x-7)^2}=\frac{1}{96(x-1)}+\frac{5}{216(x-1)^2}+\frac{1}{36(x-1)^3}+\frac{-1}{800(x+3)}+\frac{E}{(x-7)}+\frac{29}{1080(x-7)^2}$

I leave the value of the coefficient E for your homework, if the technique is of interest. If not, here is the answer

$E=\frac{-11}{1200}$

An online calculator that will do all this for you in automatic mode works, including in the field of complex numbers.

Now decomposing any correct fractional rational function into protozoa will not be the slightest difficulty.

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