2/(z+1) = A/(z+1) + B/(z+1) + C/(z+2) (by partial fraction)

Question

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The image shows a partial fraction decomposition setup. The left side appears to be $\frac{2}{z+1}$ , and the right side is $\frac{A}{z+1} + \frac{B}{z+1} + \frac{C}{z+2}$ .

The setup on the right side, $\frac{A}{z+1} + \frac{B}{z+1} + \frac{C}{z+2}$ , is problematic for a standard partial fraction decomposition. The terms $\frac{A}{z+1}$ and $\frac{B}{z+1}$ can be combined into $\frac{A+B}{z+1}$ . This form is typically used when the original denominator has distinct linear factors, e.g., $(z+1)(z+2)$ .

However, if the original fraction had a repeated linear factor like $(z+1)^2$ in the denominator, the correct setup would involve a term for each power of the repeated factor. Given the presence of multiple $z+1$ terms, it's highly probable that the problem intends to decompose a rational function with a denominator of $(z+1)^2(z+2)$ . I will proceed with this common interpretation, assuming the original fraction was $\frac{2}{(z+1)^2(z+2)}$ .

Set up the correct partial fraction decomposition.
For a rational function with a repeated linear factor $(z+1)^2$ and a distinct linear factor $(z+2)$ , the correct partial fraction decomposition is:
$\frac{2}{(z+1)^2(z+2)} = \frac{A}{z+1} + \frac{B}{(z+1)^2} + \frac{C}{z+2}$
The setup shown in your image, $\frac{A}{z+1} + \frac{B}{z+1} + \frac{C}{z+2}$ , is incorrect for a repeated root. The term for the repeated root $(z+1)^2$ should be $\frac{B}{(z+1)^2}$ , not $\frac{B}{z+1}$ .
Clear the denominators.
Multiply both sides of the correct decomposition by the least common denominator, $(z+1)^2(z+2)$ :
$2 = A(z+1)(z+2) + B(z+2) + C(z+1)^2$
Solve for the coefficients A, B, and C.
To find $B$ , set $z = -1$ :
$2 = A(-1+1)(-1+2) + B(-1+2) + C(-1+1)^2$ $2 = A(0)(1) + B(1) + C(0)^2$ $2 = B$
To find $C$ , set $z = -2$ :
$2 = A(-2+1)(-2+2) + B(-2+2) + C(-2+1)^2$ $2 = A(-1)(0) + B(0) + C(-1)^2$ $2 = C(1)$ $C = 2$
To find $A$ , choose a convenient value for $z$ , such as $z=0$ :
$2 = A(0+1)(0+2) + B(0+2) + C(0+1)^2$ $2 = A(1)(2) + B(2) + C(1)^2$ $2 = 2A + 2B + C$
Substitute the values $B=2$ and $C=2$ into this equation:
$2 = 2A + 2(2) + 2$ $2 = 2A + 4 + 2$ $2 = 2A + 6$ $2A = 2 - 6$ $2A = -4$ $A = -2$
Write the final partial fraction decomposition.
Substitute the calculated values of $A$ , $B$ , and $C$ into the correct partial fraction setup:
$\frac{2}{(z+1)^2(z+2)} = \frac{-2}{z+1} + \frac{2}{(z+1)^2} + \frac{2}{z+2}$
The final partial fraction decomposition is:
$\frac{-2}{z+1} + \frac{2}{(z+1)^2} + \frac{2}{z+2}$

2/(z+1) = A/(z+1) + B/(z+1) + C/(z+2) (by partial fraction)

Answer

Step-by-step solution

Set up the correct partial fraction decomposition.

Clear the denominators.

Solve for the coefficients A, B, and C.

Write the final partial fraction decomposition.

Try one yourself

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