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junior college 2 | H1 Maths
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How to express in partial fraction
Date Posted: 5 years ago
Views: 418
Following the partial fraction decomposition formula,
Let (2x-1) / ((x-2)²(x+1)²) = A/(x-2) + B/(x-2)² + C/(x+1) + D/(x+1)²
Multiplying both sides by the denominator of the left hand side,
2x - 1 = A(x-2)(x+1)² + B(x+1)² + C(x-2)²(x+1)
+ D(x-2)²
Sub x = 2,
2(2) - 1 = B(2 + 1)²
3 = 9B
B = 1/3
Sub x = -1,
2(-1) - 1 = D(-1 - 2)²
- 3 = 9D
D = -1/3
Sub x = 0,
-1 = -2A + B + 4C + 4D
Since D = -1/3 and B = 1/3,
-1 = -2A + 1/3 + 4C - 4/3
2A = 4C
A = 2C
Sub x = 1,
2(1) - 1 = -4A + 4B + 2C + D
1= -4A + 4/3 + A -1/3
- 3A = 0
A = 0
So since A=0, C = 0
So (2x-1) / ((x-2)²(x+1)²)
= 1/(3(x-2)²) - 1/(3(x+1)²)
Let (2x-1) / ((x-2)²(x+1)²) = A/(x-2) + B/(x-2)² + C/(x+1) + D/(x+1)²
Multiplying both sides by the denominator of the left hand side,
2x - 1 = A(x-2)(x+1)² + B(x+1)² + C(x-2)²(x+1)
+ D(x-2)²
Sub x = 2,
2(2) - 1 = B(2 + 1)²
3 = 9B
B = 1/3
Sub x = -1,
2(-1) - 1 = D(-1 - 2)²
- 3 = 9D
D = -1/3
Sub x = 0,
-1 = -2A + B + 4C + 4D
Since D = -1/3 and B = 1/3,
-1 = -2A + 1/3 + 4C - 4/3
2A = 4C
A = 2C
Sub x = 1,
2(1) - 1 = -4A + 4B + 2C + D
1= -4A + 4/3 + A -1/3
- 3A = 0
A = 0
So since A=0, C = 0
So (2x-1) / ((x-2)²(x+1)²)
= 1/(3(x-2)²) - 1/(3(x+1)²)
Wow thank you so much
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Btw you can sub random x values into both sides to check the validity of the newly expressed equation:)
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