Ex. 5. Resolve {42-19x}{(x^2 +1)(x-4)} into partial fractions.
Assume {42-19x}{(x^2 +1)(x-4)} = {Ax+B}{x^2 +1} + {C}{x-4}
42 — 19x =(Ax + B)(x-4)+ C(ax+ 1).
Let x = 4, then C= 2; equating coefficients of x^2, 0=A+C and A = 2. equating the absolute terms, 42=-4B+4C, and B=-11,
{42-19x}{(x^2 +1)(x-4)} = {2x-11}{x^2 +1} - {2}{x-4}
494. In all the preceding examples the numerator has been of lower dimensions than the denominator; if this is not the case, we divide the numerator by the denominator until a remainder is obtained which is of lower dimensions than the denominator.
Ex. Resolve {6x^3+4x^2-7}{3x^2-2x-1} into partial fractions.
By division,
{6x^3+4x^2-7}{3x^2-2x-1} = 2x+3+ {8x-4}{3x^2-2x-1}
{8x-4}{3x^2-2x-1} = {5}{3x+1} + {1}{x-1}
{6x^3+4x^2-7}{3x^2-2x-1} = 2x+3+{5}{3x+1} + {1}{x-1}
495. The General Term. We shall now explain how resolution into partial fractions may be used to facilitate the expansion of a rational fraction in ascending powers of x.
Ex. 1. Find the general term of {1+6x}{1-3x+2x^2} when expanded in a series of ascending powers of x.
By Ex. 1, Art. 493, we have
{1+5x}{1-3x+2x^2} = {7}{1-2x} - {6}{1-x} = 7(1-2x)^-1 - 6(1-x)^-1 =7[1+(2x) + (2x)^2 + + (2x)^r + ] - 6(1+x+x^2+ + x^r+ ).
Hence the (r + 1)th, or general term of the expansion, is 7(2x)^r — 6x^r or [7(2)r — 6]x^r.
