Partial Fractions: Decomposition (Type-1)


When the factors in the denominator are linear and different: Let the linear and different factors in the denominator are \((x - a), (x - b), (x - c),.....\) then write the fractions like the given below and it should be equal to the main fraction then solve the equation. $$ \frac{A}{(x - a)} + \frac{B}{(x - b)} + \frac{C}{(x - c)}.... $$ Here A, B, and C are the unknown constants.

Note: If there is a quadratic equation in the denominator, in that case, we have to factorize the quadratic equation into linear factors first. If it is not possible to factorize the quadratic equation then use the type III or Type IV method to solve the question. We can better understand by example (2) given below.

Example(1): \(\frac{1}{(x - 1) \ (x - 2)}\)

Solution: Method(1): We can see that there are two linear factors in the denominator of the given equation so we can write the equation as given below.

\(\frac{1}{(x - 1) \ (x - 2)} =\) \(\frac{A}{(x - 1)} + \frac{B}{(x - 2)}.....(1)\)

$$ \frac{1}{(x - 1) \ (x - 2)} = \frac{A}{(x - 1)} + \frac{B}{(x - 2)}.....(1) $$

Now taking the LCM of RHS (Right Hand Side) terms.

\(\frac{1}{(x - 1) \ (x - 2)} =\) \(\frac{A (x - 2) + B (x - 1)}{(x - 1) (x - 2)}\)

$$ \frac{1}{(x - 1) \ (x - 2)} = \frac{A (x - 2) + B (x - 1)}{(x - 1) (x - 2)} $$

The terms in the denominators on both sides are the same so they will be canceled. $$ 1 = A (x - 2) + B (x - 1)....(2) $$ Now to calculate the value of A, we can take the denominator of A from the equation (1) and put it equal to zero (0). $$ x - 1 = 0 $$ $$ x = 1 $$ Now put this value of \(x\) in equation (2). $$ 1 = A (1 - 2) + B (1 - 1) $$ $$ 1 = A (-1) + 0 $$ $$ A = -1 $$ Similarly to calculate the value of B, we can take the denominator of B from equation (1) and put it equal to zero (0). $$ x - 2 = 0 $$ $$ x = 2 $$ again put this value of \(x\) in equation (2). $$ 1 = A (2 - 2) + B (2 - 1) $$ $$ 1 = 0 + B (1) $$ $$ B = 1 $$ Finally, put the values of A and B in equation (1).

\(\frac{1}{(x - 1) \ (x - 2)} =\) \(\frac{-1}{(x - 1)} + \frac{1}{(x - 2)}\)

$$ \frac{1}{(x - 1) \ (x - 2)} = \frac{-1}{(x - 1)} + \frac{1}{(x - 2)} $$

Hence this is the final decomposition of the partial fraction equation.

Solution: Method(2): The given equation can also be written as

\(\frac{1}{(x - 1) \ (x - 2)} =\) \(\frac{A}{(x - 1)} + \frac{B}{(x - 2)}.....(1)\)

$$ \frac{1}{(x - 1) \ (x - 2)} = \frac{A}{(x - 1)} + \frac{B}{(x - 2)}.....(1) $$

Now taking the LCM of RHS (Right Hand Side) terms.

\(\frac{1}{(x - 1) \ (x - 2)} =\) \(\frac{A (x - 2) + B (x - 1)}{(x - 1) (x - 2)}\)

$$ \frac{1}{(x - 1) \ (x - 2)} = \frac{A (x - 2) + B (x - 1)}{(x - 1) (x - 2)} $$

The terms in the denominators on both sides are the same so they will be canceled. $$ 1 = A (x - 2) + B (x - 1)....(2) $$ Write down the terms of \(x\), and constants separately as written below. $$ 1 = x \ (A + B) + (-2A - B) $$ Now compare the values of \(x\), and constants, of RHS and LHS. $$ A + B = 0....(3) $$ $$ -2A - B = 1....(4) $$ after adding the equation (3) and (4) we get. $$ A + B - 2A - B = 0 + 1 $$ $$ A = -1 $$ now put the value of A in equation (3) to get the value of B. $$ -1 + B = 0 $$ $$ B = 1 $$ Finally, put the values of A and B in equation (1).

\(\frac{1}{(x - 1) \ (x - 2)} =\) \(\frac{-1}{(x - 1)} + \frac{1}{(x - 2)}\)

$$ \frac{1}{(x - 1) \ (x - 2)} = \frac{-1}{(x - 1)} + \frac{1}{(x - 2)} $$

Hence this is the final decomposition of the partial fraction equation.



Example(2): \(\frac{4x - 3}{3x^2 + x - 2}\)

Here the denominator is a quadratic equation so we need to factorize it into linear factors first. $$ = 3x^2 + x - 2 $$ $$ = 3x^2 + 3x - 2x - 2 $$ $$ = 3x \ (x + 1) - 2 \ (x + 1) $$ $$ = (x + 1) \ (3x - 2) $$ After putting the linear factors in the main equation we get. $$ \frac{4x - 3}{(x + 1) \ (3x - 2)} $$ Now we can solve this equation similar to the question (1).

The equation can be also written as.

\(\frac{4x - 3}{(x + 1) \ (3x - 2)} =\) \(\frac{A}{(x + 1)} + \frac{B}{(3x - 2)}.....(1)\)

$$ \frac{4x - 3}{(x + 1) \ (3x - 2)} = \frac{A}{(x + 1)} + \frac{B}{(3x - 2)}.....(1) $$

Now taking the LCM of RHS (Right Hand Side) terms.

\(\frac{4x - 3}{(x + 1) \ (3x - 2)} =\) \(\frac{A (3x - 2) + B (x + 1)}{(x + 1) (3x - 2)}\)

$$ \frac{4x - 3}{(x + 1) \ (3x - 2)} = \frac{A (3x - 2) + B (x + 1)}{(x + 1) (3x - 2)} $$

The terms in the denominator on both sides are the same so they will be canceled. $$ 4x - 3 = A (3x - 2) + B (x + 1)....(2) $$ Now to calculate the value of A, we can take denominator of A from the equation (1) and put it equal to zero (0). $$ x + 1 = 0 $$ $$ x = -1 $$ Now put this value of \(x\) in equation (2).

\(4 (-1) - 3 =\) \(A \{3 \ (-1) - 2\} + B (-1 + 1)\)

$$ 4 (-1) - 3 = A \{3 \ (-1) - 2\} + B (-1 + 1) $$

$$ - 4 - 3 = A (- 3 - 2) + 0 $$ $$ -7 = -5A $$ $$ 7 = 5A $$ $$ A = \frac{7}{5} $$ Similarly to calculate the value of B, we can take the denominator of B from equation (1) and put it equal to zero (0). $$ 3x - 2 = 0 $$ $$ 3x = 2 $$ $$ x = \frac{2}{3} $$ again put this value of \(x\) in equation (2).

\(4 \times \frac{2}{3} - 3 =\) \(A (3 \times \frac{2}{3} - 2) + B (\frac{2}{3} + 1)\)

$$ 4 \times \frac{2}{3} - 3 = A (3 \times \frac{2}{3} - 2) + B (\frac{2}{3} + 1) $$

$$ \frac{8}{3} - 3 = 0 + B (\frac{2 + 3}{3}) $$ $$ \frac{8 - 9}{3} = B \ \frac{5}{3} $$ $$ \frac{-1}{3} = \frac{5B}{3} $$ $$ 15B = -3 $$ $$ B = \frac{-3}{15} $$ Finally, put the values of A and B in equation (1).

\(\frac{4x - 3}{(x + 1) \ (3x - 2)} =\) \(\frac{7}{5 \ (x + 1)} + \frac{-3}{15 \ (3x - 2)}\)

$$ \frac{4x - 3}{(x + 1) \ (3x - 2)} = \frac{7}{5 \ (x + 1)} + \frac{-3}{15 \ (3x - 2)} $$

Hence this is the final decomposition of the partial fraction equation.

The example (2) can also be solved by method (2) as we have solved the example (1) above. Try to solve it by using the method (2) if you will face any issue, you can write us on the comment section or send the message from the contact page.




Partial Fractions

Chapter 2: Decomposition (Type-1)


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