Partial fraction decomposition

Question

Use partial fraction decomposition to solve for constants A and B:

2x+1/x^2-x-6 = A/x+2 + B/x-3

Answer 1

Note that the denominator on the left is x²-x-6, which factors out to (x+2)(x-3).

So we can write the equation as

(2x+1)/[(x+2)(x-3)] = A/(x+2) + B/(x-3).

Let's see if we can rewrite the right side of the equation in terms of the denominator on the left side.

A/(x+2) = [A(x-3)]/[(x+2)(x-3)] = (Ax-3A)/[(x+2)(x-3)]. Here we multiply the original expression by (x-3)/(x-3), which is equal to 1 and shouldn't change anything as long as x≠3.

B/(x-3) = [B(x+2)]/[(x+2)(x-3)] = (Bx+2B)/[(x+2)(x-3)]. Same as the first step, except this time we multiply by (x+2)/(x+2), which is equal to 1 as long as x≠-2.

So we can rewrite the right side of the original equation as:

A/(x+2) + B/(x-3) = [A(x-3)]/[(x+2)(x-3)] + [B(x+2)]/[(x+2)(x-3)] = (Ax-3A)/[(x+2)(x-3)] + (Bx+2B)/[(x+2)(x-3)]

Note that we now have a common denominator, so we can add the two fractions.

(Ax-3A)/[(x+2)(x-3)] + (Bx+2B)/[(x+2)(x-3)] = (Ax - 3A + Bx + 2B)/[(x+2)(x-3)].

Remember our original left-hand side? That's still equal to our current expression.

(2x+1)/[(x+2)(x-3)] = (Ax - 3A + Bx + 2B)/[(x+2)(x-3)]

Now hopefully you can see since the denominators are the same, the numerators should also be equal. You can also think of this as multiplying both sides by (x+2)(x-3).

2x + 1 = Ax-3A + Bx + 2B.

Here's the tricky part. Let's group like terms on the right-hand side together.

2x + 1 = Ax-3A + Bx + 2B = Ax + Bx - 3A + 2B = (A+B)x + (-3A + 2B).

So we have 2x + 1 = (A+B)x + (-3A + 2B).

For these two things to be equal, the corresponding coefficients need to be equal. (Consider that any real multiple of x cannot be equal to a constant, and any real multiple of 1 cannot be equal to a multiple of x).

So

1 = (-3A+2b)

and

2x = (A+B)x.

We can simplify the second equation to

2 = A + B by dividing both sides by x.

Now we have a system of equations:

1 = -3A + 2B (equation i)

2 = A + B (equation ii)

Solving the system of equations:

6 = 3A + 3B (multiply equation ii by 3)

1 = - 3A + 2B (original equation i)

7 = 5B (add the previous two steps)

7/5 = B

2 = A + 7/5 (substitution into equation ii)

A = 3/5

So A = 3/5, and B = 7/5

Answer 2

(2x+1) / [(x+2)(x-3)] = A/(x+2) + B/(x-3)

2x+1 = A(x-3) + B(x+2)

2x + 1 = (A + B)x + (-3A + 2B)

So, A + B = 2 and -3A + 2B = 1

Add 3 times the first equation to the second equation to get 5B = 7. So B = 7/5 and A = 3/5

The partial fraction decomposition is (3/5) / (x + 2) + (7/5) / (x - 3)