I'll do one. You can follow the method on the others.
Recall that
1/(x-a)^n requires terms of
A/(x-a) + B/(x-a)^2 + ... + N/(x-a)^n
Similarly for 1/(x^2+bx+c)^n. You need terms for each power, of the form
(Ax+B)/(x^2+bx+c)^k for k=1..n
3x^2/(x-1)^2 (x-2)
= A/(x-1) + B/(x-1)^2 + C/(x-2)
clearing the fractions and ignoring the common denominator,
3x^2 = A(x-1)(x-2) + B(x-2) + C(x-1)^2
3x^2 = (A+C)x^2+(-3A+B-2C)x+2A-2B+C
For the two polynomials to be identical, the coefficients must match
A+C = 3
-3A+B-2C = 0
2A-2B+C = 0
A = -9
B = -3
C = 12
3x^2/(x-1)^2(x-2) = -9/(x-1) - 3/(x-1)^2 + 12/(x-2)
Separate the following into partial fractions:
(a) 3x^2/(x-1)^2 (x-2)
(b) x-1/(x^2 + 2)(x + 1)
(c) x^2 + 1/(x + 1)(x - 3)
1 answer
2 answers