Question

Resolve into partial fractions

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Answer 1

See below

Explanation:

1)

`(11 + x)/((2 - x)(x - 3)) = (A)/(2 - x) + (B)/(x - 3) \\ \\ \therefore \: (11 + x)/((2 - x)(x - 3)) = (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 \\ \\ (11 + x)/((2 - x)(x - 3)) = ( - 13)/(2 - x) + ( - 14)/(x - 3) \\ \\ (11 + x)/((2 - x)(x - 3)) = - ( 13)/(2 - x) - ( 14)/(x - 3)`

2)

`(12x + 11)/(x^2 +x - 6)`

`=(12x + 11)/(x^2 +3x-2x - 6)`

`=(12x + 11)/(x(x +3) -2(x +3))`

`\therefore (12x + 11)/(x^2 +x - 6)=(12x + 11)/((x +3) (x - 2))`

`\therefore (12x + 11)/(x^2 +x - 6)=(A)/((x +3))+(B)/((x - 2))`

`\therefore (12x + 11)/(x^2 +x - 6)=(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 (12x + 11)/(x^2 +x - 6)=(5)/((x +3))+(7)/((x - 2))`