Video: Partial Fraction Decomposition

Find 𝐴 and 𝐡 such that 2π‘₯/(π‘₯ βˆ’ 3)Β² = 𝐴/(π‘₯ βˆ’ 3) + 𝐡/(π‘₯ βˆ’ 3)Β².

04:48

Video Transcript

Find 𝐴 and 𝐡 such that two π‘₯ over π‘₯ minus three squared is equal to 𝐴 over π‘₯ minus three plus 𝐡 over π‘₯ minus three squared.

In order to answer this question, we need to recall how we split a term in the form 𝑝π‘₯ plus π‘ž over π‘₯ minus π‘Ž squared into partial fractions. This term can be written in the form 𝐴 over π‘₯ minus π‘Ž plus 𝐡 over π‘₯ minus π‘Ž squared as shown in the question. The value of 𝑝 in our question is two, π‘ž is equal to zero, and lowercase π‘Ž is equal to three.

We begin by multiplying all three of our terms by π‘₯ minus three squared. At this point, we can cancel common terms on the numerator and denominator. The left-hand side cancels to become two π‘₯. On the first term on the right-hand side, we can cancel π‘₯ minus three, leaving us 𝐴 multiplied by π‘₯ minus three. The final term is 𝐡 as the π‘₯ minus three all squared cancel. Our equation is, therefore, two π‘₯ equals 𝐴 multiplied by π‘₯ minus three plus 𝐡.

There are two traditional methods for calculating 𝐴 and 𝐡 from this point: firstly, by comparing coefficients and secondly, by substitution. Let’s firstly solve the equation by comparing coefficients. Distributing the parentheses or expanding the brackets on the right-hand side gives us 𝐴π‘₯ minus three 𝐴. This gives us two π‘₯ is equal to 𝐴π‘₯ minus three 𝐴 plus 𝐡. We can now consider the π‘₯ terms on both sides of the equation and compare the coefficients. On the left-hand side, we have two π‘₯ and on the right-hand side, 𝐴π‘₯. Therefore, two is equal to 𝐴. We noticed that there is no constant on the left-hand side. But on the right-hand side, we have negative three 𝐴 plus 𝐡. This means that zero is equal to negative three 𝐴 plus 𝐡.

Substituting in our value for 𝐴 gives us zero is equal to negative three multiplied by two plus 𝐡. Negative three multiplied by two is equal to negative six. Adding six to both sides of this equation gives us six is equal to 𝐡. Using partial fractions and then comparing coefficients, we have found that our values for 𝐴 and 𝐡 are two and six, respectively.

We’ll now look at the second method involving substitution. We return to the equation two π‘₯ is equal to 𝐴 multiplied by π‘₯ minus three plus 𝐡. We can substitute any two values into this equation to calculate 𝐴 and 𝐡. However, if we substitute π‘₯ equals three, this will eliminate the parentheses. On the left-hand side, we have two multiplied by three and on the right-hand side, 𝐴 multiplied by three minus three plus 𝐡. As three minus three is equal to zero, the 𝐴 term cancels. Two times three is equal to six. Therefore, once again, we have the solution 𝐡 is equal to six.

As π‘₯ minus three was the only parentheses, we can now choose any integer to substitute into the equation. It is usually sensible to choose zero or one. If π‘₯ is equal to zero, the left-hand side will equal zero. On the right-hand side, we have 𝐴 multiplied by zero minus three plus 𝐡. This simplifies to zero is equal to negative three 𝐴 plus 𝐡. As 𝐡 is equal to six, we can substitute in this value. Adding three 𝐴 to both sides gives us three 𝐴 is equal to six. Finally, dividing both sides by three gives us 𝐴 is equal to two. Once again, this gives us the same answer as our first method.

The answers to this question are 𝐴 is equal to two and 𝐡 is equal to six.

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