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How would you express (11+x-x^2)/[(x+1)(x-2)^2] in terms of partial fractions?

The first thing to consider when converting an expression to impartial fractions is how many fractions you will have in the end. With linear (no powers) brackets, found in the denominator of your initial fraction, the number of partial fractions at the end of your workings coincides with the number of said brackets. For example, a fraction with denominator (x+5)(x-2) would mean a final answer consisting of 2 fractions. However, in cases involving powers, such as this one, there will be an additional fraction in the answer. So, in this case, (x+1)(x-2)^2 will mean an answer consisting of 3 brackets.

            Next, we write our fraction out into a kind of placeholder, with A, B and C used in our numerators to represent the answers we hope to calculate. The denominators will be our original brackets on the denominator of the question, each separated into their own fraction, as follows:

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

Notice that the B and C fractions are almost identical apart from the power of 2 added to only one of the brackets. This power, as well as the linear form in B, ensures we have partial fractions that, if added together once again, would create the original expression in the question.

Next, we multiply through by the original denominator, the brackets. This will create the equation [1] as follows:

[1]   11+x-x^2 = A(x-2)^2 + B(x+1)(x-2) + C(x+1)

An easy way to work out what your expression should look like in this step is to compare what you have in your brackets through which you are multiplying, and then to look at what is in the denominator of each fraction. Each term (A,B,C etc) will essentially be multiplied by what it does not have in its denominator, as the bracket it does have will cancel. We will use A in the long-hand way as an example:

[A/(x+1)] * [(x+1)(x-2)^2] = [A(x+1)(x-2)^2)]/(x+1) = A(x-2)^2

In this example, the two (x+1) brackets, top and bottom, cancelled, leaving us with the answer. The next step is to look at the brackets paired with our letters, and determine an x value that will equal as many terms as possible to zero. Lets take x=-1 , and substitute this into equation [1]. This will mean our B and C fractions will equal to zero, as follows:

11+(-1)-(-1)^2 = A(-1-2)^2 + B(0) + C(0) = A(9)

We can now rearrange this to find A:

9 = 9A

A = 1

We then repeat this method with new x values to find the value of C:

x=2

11+2-4 = A(0) + B(0) + C(3)

3C = 9

C = 3

To find B, we can substitute in our values for A and C, as well as any x value (other than those already used):

x=1

11+1-1 = 1(-1)^2 + B(2)(-1) + 3(2)

-2B = 11-1-6

B = 4/-2 = -2

To complete our answer, we rewrite equation [1], with our values for A, B and C substituted in:

(11+x-x^2)/[(x+1)(x-2)^2] = 1/(x+1) – 2/(x-2) + 3/(x-2)^2

This is our answer, in partial fractions.

Matthew M. A Level Maths tutor, GCSE Maths tutor, GCSE History tutor,...

10 months ago

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