How do you write the partial fraction decomposition of the rational expression $(5x - 1) / ((x - 2)(x + 1))$ ?

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Answer 1 · 2016-01-06T01:02:47

$(5x-1)/((x-2)(x+1)) = 3/(x-2) + 2/(x+1)$

The partial fraction decomposition suggests that the function can be broken down into the sum of two other functions, or;

$(5x-1)/((x-2)(x+1)) = A/(x-2) + B/(x+1)$

Where we need to solve for $A$ and $B$ . We can cross multiply to combine the terms on the right hand side over a common denominator. We get;

$(5x-1)/((x-2)(x+1)) = (A(x+1) + B(x-2))/((x-2)(x+1))$

We can now cancel the denominator on each side, leaving;

$5x-1 = A(x+1) + B(x-2)$

Now we can solve for $A$ and $B$ . We can make one of the terms cancel out by choosing the right value for $x$ . Lets try $x=~1$ .

$5(~1)-1 = A(~1+1) + B(~1-2)$

The $A$ term goes away since it is multiplied by zero, leaving;

$~6 = ~3B$

Solving for $B$ ;

$B=2$

We can substitute $B$ and solve for $A$ , but it would be easier to do the same trick that we used to solve for $B$ . Let $x=2$ .

$5(2) -1 = A(2+1) + B(2-2)$

This time, the $B$ term goes away;

$9 = 3A$

$A = 3$

Now that we have our values for $A$ and $B$ we can plug into our first function and get;

$(5x-1)/((x-2)(x+1)) = 3/(x-2) + 2/(x+1)$

How do you write the partial fraction decomposition of the rational expression (5x - 1) / ((x - 2)(x + 1))(5x - 1) / ((x - 2)(x + 1))?