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Chapter 16: Problem 6

Resolve \(\frac{2}{x\left(x^{2}-3 x+2\right)}\) into partial fractions. a) \(\frac{1}{x}+\frac{1}{x-2}-\frac{2}{x-1}\) c) \(\frac{1}{x}-\frac{2}{x-2}+\frac{1}{x-1}\) b) \(\frac{1}{x}-\frac{2}{x-2}+\frac{1}{x-1}\) d) \(\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x-1}\)

Short Answer

Expert verified
Question: Determine the partial fraction decomposition of the given expression: \(\frac{2}{x(x^2-3x+2)}\) a) \(\frac{2}{x}-\frac{1}{x-1}+\frac{1}{x-2}\) b) \(\frac{1}{x}+\frac{2}{x-1}-\frac{1}{x-2}\) c) \(\frac{1}{x}-\frac{2}{x-2}+\frac{1}{x-1}\) d) \(\frac{1}{x}+\frac{1}{x-1}+\frac{2}{x-2}\) Answer: c) \(\frac{1}{x}-\frac{2}{x-2}+\frac{1}{x-1}\)

Step by step solution

01

Identify the factors of the denominator

The given fraction is \(\frac{2}{x(x^2-3x+2)}\). We need to factor the quadratic expression: \(x^2 - 3x + 2\) which can be factored as \((x-1)(x-2)\). Now, we have the expression: \(\frac{2}{x(x-1)(x-2)}\).
02

Set up partial fractions

According to the partial fractions theorem, the given expression can be written as the sum of partial fractions with the factors of the denominator: \(\frac{2}{x(x-1)(x-2)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x-2}\)
03

Clear the denominators

To find the constants A, B, and C, we need to clear the denominators by multiplying both sides by \(x(x-1)(x-2)\): \(2 = A(x-1)(x-2) + Bx(x-2) + Cx(x-1)\)
04

Find the constants A, B, and C

We can find A, B, and C by substituting appropriate values for x: For A: Set x=0, so everything except the term containing A becomes zero: \(2 = A(-1)(-2) \Rightarrow A = \frac{2}{2}=1\) For B: Set x=1, so everything except the term containing B becomes zero: \(2 = B(1)(1-2) \Rightarrow B = \frac{2}{-1} = -2\) For C: Set x=2, so everything except the term containing C becomes zero: \(2 = C(2)(2-1) \Rightarrow C = \frac{2}{2} = 1\)
05

Write the final expression

Now that we have the constants A, B, and C, we can rewrite the original expression as the sum of partial fractions: \(\frac{2}{x(x-1)(x-2)} = \frac{1}{x}-\frac{2}{x-2}+\frac{1}{x-1}\) So, the answer is c) \(\frac{1}{x}-\frac{2}{x-2}+\frac{1}{x-1}\).

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