Question

Resolve into partial fractions​

Answer 1

Answer:

See below

Step-by-step explanation:

1)

$\frac{11 + x}{(2 - x)(x - 3)} = \frac{A}{2 - x} + \frac{B}{x - 3} \\ \\ \therefore \: \frac{11 + x}{(2 - x)(x - 3)} = \frac{A(x - 3) +B(2 - x) }{(2 - x)(x - 3)} \\ \\ \therefore \: 11 + x = Ax - 3A + 2B - Bx \\ \\ \therefore \: 11 + x = - 3A + 2B + Ax - Bx\\ \\ \therefore \: 11 + x = - 3A + 2B + (A - B)x \\ \\ equating \: the \: like \: terms \: on \: both \: sides \\ A - B = 1 \\ \therefore \: A = B + 1.....(1) \\ \\ - 3A + 2B = 11....(2) \\ from \: eq \: (1) \: and \: (2) \\ - 3(B + 1) + + 2B = 11 \\ \\ - 3B - 3 + 2B = 11 \\ \\ - B = 11 + 3 \\ \\ B = - 14 \\ A = - 14 + 1 = - 13 \\ \\ \frac{11 + x}{(2 - x)(x - 3)} = \frac{ - 13}{2 - x} + \frac{ - 14}{x - 3} \\ \\ \frac{11 + x}{(2 - x)(x - 3)} = - \frac{ 13}{2 - x} - \frac{ 14}{x - 3}$

2)

$\frac{12x + 11}{x^2 +x - 6}$

$=\frac{12x + 11}{x^2 +3x-2x - 6}$

$=\frac{12x + 11}{x(x +3) -2(x +3)}$

$\therefore \frac{12x + 11}{x^2 +x - 6}=\frac{12x + 11}{(x +3) (x - 2)}$

$\therefore \frac{12x + 11}{x^2 +x - 6}=\frac{A}{(x +3)}+\frac{B}{(x - 2)}$

$\therefore \frac{12x + 11}{x^2 +x - 6}=\frac{A(x-2)+B(x+3)}{(x-2)(x+3)}$

$\therefore 12x + 11 = A(x-2)+B(x+3)$

$\therefore 12x + 11 = Ax-2A+Bx+3B$

$\therefore 12x + 11 = Ax+Bx-2A+3B$

$\therefore 12x + 11 = (A+B) x-2A+3B$

Equating like terms on both sides:

$A + B = 12\implies A = 12-B... (1)$

$- 2A + 3B = 11... (2)$

Solving equations (1) & (2), we find:

A = 5, B = 7

$\therefore \frac{12x + 11}{x^2 +x - 6}=\frac{5}{(x +3)}+\frac{7}{(x - 2)}$