partial fractions

Question

Consider the partial fraction decomposition of the following

1/(x-3)(x-7)^2

One of the terms in the partial fraction decomposition is of the form:

B/(x-7)

Find the value of B.

Answer 1

This allows you to write three linear equations. Since the numerator of our expression (1/[(x-3)(x-7)²] contains NO x² terms, then

(1) A + B = 0

Since the numerator of our expression (1/[(x-3)(x-7)²] contains NO "x" terms, then

(2) -14A - 10B + C = 0

And since the constant term of our expression (1/[(x-3)(x-7)²] is "1" then:

(3) 49A + 21B - 3C = 1

Solving simultaneously A = 1/16, B = -1/16 and C = 1/4

So B = -1/16

Answer 2

Hi Sienna,

Let's start by rewriting the given expression as a partial fraction decomposition:

1/[(x-3)(x-7)^2] = A/(x-3) + B/(x-7) + C/(x-7)^2

Now we multiply both sides of the equation by the denominator (x-3)(x-7)^2 to eliminate the denominators on both sides of the equation:

1 = A(x-7)^2 + B(x-3)(x-7) + C(x-3)

Now, let's expand terms:

1 = A(x^2 - 14x + 49) + B(x^2 - 10x + 21) + C(x-3)

Next, we can expand and collect like terms:

1 = (A + B)x^2 + (-14A - 10B + C)x + (49A + 21B - 3C).

Notice that there are no x^2 terms on the left, so the coefficient (A+B) on the right side must be 0. There are also no x terms on the left, so the (-14A - 10B + C) term must be 0. Finally, since 1 is on the left, it must be equal to (49A + 21B - 3C). Let's write everything out nicely to see where we stand:

A+B=0

-14A - 10B + C = 0

49A + 21B - 3C = 1.

We can now proceed by eliminating terms. We can add 3 times the second equation to the third equation to eliminate C:

A+B = 0

7A - 9B = 1 (again, this was from adding 3 times the second eqn to the third)

Because we want to solve for B, we can subtract 7 times the first equation from the second to eliminate A and leave B:

-16B = 1.

Therefore B = -1/16.

Hope this helps!