# Video: Partial Fraction Decomposition

Find π΄ and π΅ such that 2π₯/(π₯ β 3)Β² = π΄/(π₯ β 3) + π΅/(π₯ β 3)Β².

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### Video Transcript

Find π΄ and π΅ such that two π₯ over π₯ minus three squared is equal to π΄ over π₯ minus three plus π΅ over π₯ minus three squared.

In order to answer this question, we need to recall how we split a term in the form ππ₯ plus π over π₯ minus π squared into partial fractions. This term can be written in the form π΄ over π₯ minus π plus π΅ over π₯ minus π squared as shown in the question. The value of π in our question is two, π is equal to zero, and lowercase π 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 π΄ multiplied by π₯ minus three. The final term is π΅ as the π₯ minus three all squared cancel. Our equation is, therefore, two π₯ equals π΄ multiplied by π₯ minus three plus π΅.

There are two traditional methods for calculating π΄ and π΅ from this point: firstly, by comparing coefficients and secondly, by substitution. Letβs firstly solve the equation by comparing coefficients. Distributing the parentheses or expanding the brackets on the right-hand side gives us π΄π₯ minus three π΄. This gives us two π₯ is equal to π΄π₯ minus three π΄ plus π΅. 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, π΄π₯. Therefore, two is equal to π΄. We noticed that there is no constant on the left-hand side. But on the right-hand side, we have negative three π΄ plus π΅. This means that zero is equal to negative three π΄ plus π΅.

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

Weβll now look at the second method involving substitution. We return to the equation two π₯ is equal to π΄ multiplied by π₯ minus three plus π΅. We can substitute any two values into this equation to calculate π΄ and π΅. 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, π΄ multiplied by three minus three plus π΅. As three minus three is equal to zero, the π΄ term cancels. Two times three is equal to six. Therefore, once again, we have the solution π΅ 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 π΄ multiplied by zero minus three plus π΅. This simplifies to zero is equal to negative three π΄ plus π΅. As π΅ is equal to six, we can substitute in this value. Adding three π΄ to both sides gives us three π΄ is equal to six. Finally, dividing both sides by three gives us π΄ is equal to two. Once again, this gives us the same answer as our first method.

The answers to this question are π΄ is equal to two and π΅ is equal to six.