How do you express as a partial fraction $(5x-1)/(x^2-x-2)$ ?

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Answer 1 · 2015-07-18T16:05:34

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

$(5x-1)/(x^2-x-2)$

The denominator can be factored using real numbers, so we do that first:

$x^2-x-2 = (x-2)(x+1)$

Now we want:

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

One way of proceeding is to get a single fraction on the left:

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

$= ((Ax+Bx)+(A-2B))/(x^2-x-2)$

So we want:

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

That means we need:

$A+B$ $" " =$ $" "$ $5$
$A-2B$ $" "$ $=$ $-1$

Subtract the second from the first (change the signs and add) to get:

$3B$ $=$ $6$

So $B=2$ and in order to get $A+B=5$ we must also have $A=3$

We have:

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

It is worth taking a moment to check our answer:

$(overbrace(3(x+1))^(3x+3)+overbrace(2(x-2))^(2x-4))/((x-2)(x+1))$ .

Yes the top simplifies to $5x-1$ , so we should be OK.

How do you express as a partial fraction (5x-1)/(x^2-x-2)(5x-1)/(x^2-x-2)?