⑴ Theoretically, it can be proved that any integer coefficient polynomial f(x) can be decomposed into the product of several linear factors and quadratic factors. Note that there may be the same primary factor and the same secondary factor, and the factorization of f(x) should be written in a general form.
f(x)= 1
a 1x+b 1)^(p 1(a2x+b2)^(p2)……(aix+bi)^(pi)
(c 1x^2+d 1x+e 1)^(q 1)(c2x^2+d2x+e2)^(q2)…
(cjx^2+djx+ej)^(qj).①
Where p 1, p2, …, pi, q 1, q2, …, qj are integers greater than or equal to zero.
⑵ For the partial fraction of rational function g(x)/f(x), it can be assumed that the degree of molecular polynomial is less than that of denominator polynomial (called "rational proper fraction"). (When the numerator degree is greater than or equal to the denominator degree, it can always be converted into polynomial+rational true fraction by division with remainder).
As we all know, the sum of the simplest fraction is always set by the undetermined coefficient method of the partial fraction of rational true fraction g(x)/f(x), and then the equation is obtained by comparing the coefficients of the same power of the molecule X on both sides of the equation, and then the undetermined coefficient is obtained by solving the equation to complete the partial fraction. Only when setting g(x)/f(x) as the simplest score, two points should be noted:
First, if the denominator f(x) contains the factorized factor (ax+b) k, then the set g(x)/f(x) should contain the sum of k simplest fractions:
a 1/(ax+b)+a2/(ax+b)^2+…+ak/(ax+b)^k，
Not just one AK/(ax+b) k!
Second, similar to the previous point, if f(x) contains quadratic factor (AX 2+BX+C) K after factorization, then g(x)/f(x) should contain k simplest fraction:
(b 1x+c 1)/(ax^2+bx+c)+(b2x+c2)/(ax^2+bx+c)^2+…+(bkx+ck)/(ax^2+bx+c)^k，
Instead of only items with the denominator of (ax 2+bx+c) k!