Algebra

Partial Fractions

In the resolution of a fraction into partial fractions four cases present themselves, which are illustrated in the following examples.

Example

First: When there are no repeated factors in the denominator of the given fraction. To resolve 3𝑥−2(𝑥−1)(𝑥−2)(𝑥−3) into partial fractions. Assume 3𝑥−2(𝑥−1)(𝑥−2)(𝑥−3)=𝐴(𝑥−1)+𝐵(𝑥−2)+𝐶(𝑥−3)  =𝐴(𝑥−2)(𝑥−3)(𝑥−1)(𝑥−2)(𝑥−3)+𝐵(𝑥−1)(𝑥−3)(𝑥−1)(𝑥−2)(𝑥−3)+𝐶(𝑥−1)(𝑥−2)(𝑥−1)(𝑥−2)(𝑥−3)  =𝐴(𝑥−2)(𝑥−3)+𝐵(𝑥−1)(𝑥−3)+𝐶(𝑥−1)(𝑥−2)(𝑥−1)(𝑥−2)(𝑥−3) ∴ 3𝑥−2=𝐴(𝑥−2)(𝑥−3)+𝐵(𝑥−1)(𝑥−3)+𝐶(𝑥−1)(𝑥−2) Since 𝐴,𝐵, and 𝐶 do not contain 𝑥, and this equation is true for all values of 𝑥, put 𝑥=1; then 3−2=𝐴(1−2)(1−3), from which 𝐴=12 Similarly, if 𝑥 be put =2, we have 6−2=𝐵(2−1)(2−3), ∴ 𝐵=−4 and, putting 𝑥=3, 9−2=𝐶(𝑥−1)(𝑥−2), ∴ 𝐶=72 Hence 3𝑥−2(𝑥−1)(𝑥−2)(𝑥−3)=12(𝑥−1)−4(𝑥−2)+72(𝑥−3) 235

Example

Secondly. When there is a repeated factor. Resolve into partial fractions 7𝑥³−10𝑥²+6𝑥(𝑥−1)³(𝑥+2) Assume 7𝑥³−10𝑥²+6𝑥(𝑥−1)³(𝑥+2)=𝐴(𝑥−1)³+𝐵(𝑥−1)²+𝐶𝑥−1+𝐷𝑥+2 These forms are necessary and sufficient. Multiplying up, we 7𝑥3−10𝑥2+6𝑥=𝐴(𝑥+2)+𝐵(𝑥−1)(𝑥+2)+𝐶(𝑥−1)2(𝑥+2)+𝐷(𝑥−1)31 Make 𝑥=1; ∴ 7−10+6=𝐴(1+2); ∴ 𝐴=1 Substitute this value of 𝐴 in [1]; thus 7𝑥3−10𝑥2+6𝑥=𝑥+2+𝐵(𝑥−1)(𝑥+2)+𝐶(𝑥−1)2(𝑥+2)+𝐷(𝑥−1)3 7𝑥3−10𝑥2+5𝑥-2=𝐵(𝑥−1)(𝑥+2)+𝐶(𝑥−1)2(𝑥+2)+𝐷(𝑥−1)3 Divide by 𝑥−1; thus 7𝑥2−3𝑥+2=𝐵(𝑥+2)+𝐶(𝑥−1)(𝑥+2)+𝐷(𝑥−1)22 Make 𝑥=1 again, 7−3+2=𝐵(1+2); ∴ 𝐵=2 Substitute this value of 𝐵 in [2], and we have 7𝑥²−5𝑥−2=𝐶(𝑥−1)(𝑥+2)+𝐷(𝑥−1)² Divide by 𝑥−1; 7𝑥+2=𝐶(𝑥+2)+𝐷(𝑥−1)3 Put 𝑥=1 a third time, 7+2=𝐶(1+2); ∴ 𝐶=3 Lastly, make 𝑥=−2 in [3], −14+2=𝐷(−2−1); ∴ 𝐷=4 Result 1(𝑥−1)³+2(𝑥−1)²+3𝑥−1+4𝑥+2 236

Example

Thirdly. when there is a quadratic factor of imaginary roots not repeated. Resolve 1(𝑥²+1)(𝑥²+𝑥+1) into partial fractions. Here we must assume 1(𝑥²+1)(𝑥²+𝑥+1)=𝐴𝑥+𝐵𝑥²+1+𝐶𝑥+𝐷𝑥²+𝑥+1 𝑥²+1 and 𝑥²+𝑥+1 have no real factors, and are therefore retained as denominators. The requisite form of the numerators is seen by adding together two simple fractions, such as 𝑎𝑥+𝑏+𝑐𝑥+𝑑. Multiplying up, we have the equation 1=(𝐴𝑥+𝐵)(𝑥2+𝑥+1)+(𝐶𝑥+𝐷)(𝑥2+1)1 Let 𝑥²+1=0; ∴ 𝑥²=−1 Substitute this value of 𝑥² in [1] repeatedly; thus 1=(𝐴𝑥+𝐵)𝑥=𝐴𝑥²+𝐵𝑥=−𝐴+𝐵𝑥 or 𝐵𝑥−𝐴−1=0 Equate coefficients to zero; ∴ 𝐵=0, 𝐴=−1 Again, let 𝑥²+𝑥+1=0; ∴ 𝑥²=−𝑥−1 Substitute this value of 𝑥² repeatedly in [1]; thus 1=(𝐶𝑥+𝐷)(−𝑥)=−𝐶𝑥²−𝐷𝑥=𝐶𝑥+𝐶−𝐷𝑥; (𝐶−𝐷)𝑥+𝐶−1=0 Equate coefficients to zero; thus 𝐶=1, 𝐷=1. Hence 1(𝑥²+1)(𝑥²+𝑥+1)=𝑥𝑥²+1+𝑥+1𝑥²+𝑥+1 237

Example

Fourthly, when there is a repeated quadratic factor of imaginary roots. Resolve 40𝑥−103(𝑥+1)²(𝑥²−4𝑥+8)³ into partial fractions. Assume 40𝑥−103(𝑥+1)2(𝑥2−4𝑥+8)3=𝐴𝑥+𝐵(𝑥2−4𝑥+8)3+𝐶𝑥+𝐷(𝑥2−4𝑥+8)2+𝐸𝑥+𝐹𝑥2−4𝑥+8+𝐺(𝑥+1)2+𝐻𝑥+1 ∴ 40𝑥−103={(𝐴𝑥+𝐵)+(𝐶𝑥+𝐷)(𝑥2−4𝑥+8)+(𝐸𝑥+𝐹)(𝑥2−4𝑥+8)2}(𝑥+1)2+{𝐺+𝐻(𝑥+1)}(𝑥2−4𝑥+8)31 In the first place, to determine 𝐴 and 𝐵, equate 𝑥²−4𝑥+8 to zero; thus 𝑥²=4𝑥−8 Substitute this value of 𝑥² repeatedly in [1], as in the previous example, until the first power of 𝑥 alone remains. The resulting equation is 40𝑥−103=(17𝐴+6𝐵)𝑥−48𝐴−7𝐵 Equating coefficients, we obtain two equations 17𝐴+6𝐵=4048𝐴+7𝐵=103}, from which 𝐴=2𝐵=1 Next, to determine 𝐶 and 𝐷, substitute these values of 𝐴 and 𝐵 in [1]; the equation will then be divisible by 𝑥²−4𝑥+8. Divide, and the resulting equation is 0=2𝑥+13+{𝐶𝑥+𝐷+(𝐸𝑥+𝐹)(𝑥2−4𝑥+8)}(𝑥+1)2+{𝐺+𝐻(𝑥+1)}(𝑥2−4𝑥+8)22 Equate 𝑥²−4𝑥+8 again to zero, and proceed exactly as before, when finding 𝐴 and 𝐵. Next, to determine 𝐸 and 𝐹, substitute the values of 𝐶 and 𝐷, last found in equation [2]; divide, and proceed as before. Lastly, 𝐺 and 𝐻 are determined by equating 𝑥+1 to zero successively, as in Example 2. 238

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive