Video: Partial Fraction Decomposition

Find 𝐴 and 𝐡 such that (4π‘₯ βˆ’ 2)/(π‘₯ + 3)(π‘₯ βˆ’ 2) = 𝐴/(π‘₯ + 3) + 𝐡/(π‘₯ βˆ’ 2).

03:26

Video Transcript

Find 𝐴 and 𝐡 such that four π‘₯ minus two over π‘₯ plus three times π‘₯ minus two equals 𝐴 over π‘₯ plus three plus 𝐡 over π‘₯ minus two.

We have this equation that we want to be true for all values of π‘₯, and we need to find 𝐴 and 𝐡 to make this so. Let’s get rid of all these fractions by multiplying up by π‘₯ plus three times π‘₯ minus two.

So on the left-hand side, you have four π‘₯ minus two, and on the right-hand side, we have 𝐴 times π‘₯ plus three times π‘₯ minus two over π‘₯ plus three plus 𝐡 times π‘₯ plus three times π‘₯ minus two over π‘₯ minus two. And we can see that there’s some cancelation that will happen.

The factor of π‘₯ plus three in the numerator of the first fraction cancels with the denominator, and the same is true in the second fraction, where the factor of π‘₯ minus two cancels with the denominator. Tidying up a little, we still have four π‘₯ minus two on the left-hand side and we have 𝐴 times π‘₯ minus two plus 𝐡 times π‘₯ plus three on the right-hand side, and we can expand the brackets on the right-hand sides to get 𝐴π‘₯ minus two 𝐴 plus 𝐡π‘₯ plus three 𝐡, and we can combine like terms to get 𝐴 plus 𝐡 times π‘₯ plus negative two 𝐴 plus three 𝐡.

The reason that we needed to combine the like terms on the right-hand side was to allow us to compare coefficients. On the left-hand side, the coefficient of π‘₯ is four, and on the right-hand side, the coefficient of π‘₯ is 𝐴 plus 𝐡, and these coefficients must be equal, so four must be equal to 𝐴 plus 𝐡.

For a very similar reason, the constant term on the left, negative two, must be equal to the constant term on the right, negative two 𝐴 plus three 𝐡. Having gone through all this algebra, we’re left with two linear equations in the variables 𝐴 and 𝐡. And to finish, we just need to solve these equations simultaneously to obtain the values of 𝐴 and 𝐡.

There are a number of ways of solving simultaneous equations. We could rearrange the first equation to get 𝐡 in terms of 𝐴 and then substitute this into our second equation to get an equation in 𝐴 alone, which we can then solve in the normal way, multiplying out the brackets, combining like terms, and rearranging to find that 𝐴 is 14 over five.

Given the value of 𝐴, we can substitute this into our expression for 𝐡 in terms of 𝐴 to get that 𝐡 is four minus 14 over five, which is six over five. There are of course other methods for solving these simultaneous equations, for example, elimination or matrix methods, but all of them will give the same answer that 𝐴 is 14 over five and 𝐡 is six over five.

In the context of our question, this means that, for all π‘₯, four π‘₯ minus two over π‘₯ plus three times π‘₯ minus two is equal to 14 over five over π‘₯ plus three plus six over five over π‘₯ minus two.

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