### Video Transcript

The expression three π₯ minus two over π₯ squared plus four multiplied by π₯ minus three can be written in the form π΄π₯ plus π΅ over π₯ squared plus four plus πΆ over π₯ minus three. Find the values of π΄, π΅, and πΆ.

What weβve been asked to find here is the partial fractions decomposition of this expression. Weβre taking it from a single expression into the sum of two fractions whose denominators are the factors of the original denominator, π₯ squared plus four and π₯ minus three. To do this, letβs first consider how we would go the other way. So how would we combine the two fractions on the right to a single fraction with a common denominator?

Well, for the first fraction, we need to multiply its denominator by π₯ minus three. And so, in order for this to be equivalent, we need to do the same in the numerator. For the second fraction, weβd need to multiply its denominator by π₯ squared plus four. So again, weβd have to do the same in the numerator. Doing so would give the right-hand side as π΄π₯ plus π΅ multiplied by π₯ minus three over π₯ squared plus four multiplied by π₯ minus three plus πΆ multiplied by π₯ squared plus four over π₯ minus three multiplied by π₯ squared plus four. And now, we have a common denominator for these two terms. So we could write them as a single fraction.

Now, the key point is that the fraction on the left and the fraction on the right now have exactly the same denominators. So if these two fractions are equal to one another, it must be the case that the numerators are equal to one another. So we can rewrite our equation using the numerators only, and we have a simplified equation. Three π₯ minus two equals π΄π₯ plus π΅ multiplied by π₯ minus three plus πΆ multiplied by π₯ squared plus four.

Now, weβre looking to find the values of π΄, π΅, and πΆ. And to do this, we remember that this equation is true for all values of π₯. So the first approach we can try is to choose values of π₯ which will eliminate some of these constants. If we choose, for example, π₯ equals three, then π₯ minus three will be equal to zero. Which means weβll be able to eliminate the first part of the expression on the right-hand side. On the left-hand side, weβll have three multiplied by three minus two. And on the right-hand side, weβll be left with πΆ multiplied by three squared plus four. This gives nine minus two equals πΆ multiplied by nine plus four or seven equals 13πΆ. And dividing through by 13, we see that πΆ is equal to seven over 13.

So weβve found one of our three values. We then consider whether itβll be possible to choose an π₯-value that makes the second term on the right-hand side equal to zero. Well, in fact, this isnβt possible because it would be required that π₯ squared is equal to negative four. And this isnβt the case for any real values of π₯. So we wonβt be able to take this approach. What we can do instead then is consider a method called equating coefficients. We look at the coefficients of, for example, π₯ squared, π₯, and the constant term on the two sides of our equation. Before we can do this, though, weβll need to distribute the parentheses on the right-hand side.

Doing so gives π΄π₯ multiplied by π₯ which is π΄π₯ squared, π΄π₯ multiplied by negative three which is negative three π΄π₯, π΅ multiplied by π₯ which is π΅π₯, and π΅ multiplied by negative three which is negative three π΅. In the second set, we have πΆ multiplied by π₯ squared which is πΆπ₯ squared and πΆ multiplied by four which is four πΆ. Letβs now consider the coefficients of π₯ squared first of all. On the left-hand side, there are no π₯ squareds. So the coefficient is zero. And on the right-hand side, we have π΄ multiplied by π₯ squared plus πΆ multiplied by π₯ squared. So the coefficient of π₯ squared on the right-hand side is π΄ plus πΆ.

Now, we know the value of πΆ. Weβve already found itβs seven over 13. So we can substitute this value of πΆ in order to give an equation we can solve for π΄. We have zero equals π΄ plus seven over 13. And subtracting seven over 13 from each side, we find that the value of π΄ is negative seven over 13. To find the value of π΅, we can consider either the coefficient of π₯ or the value of the constant term because we have both π΅π₯ and negative three π΅ on the right-hand side. And the choice is somewhat arbitrary. Letβs consider then the constant term. On the left-hand side, the constant term is negative two. On the right-hand side, we have negative three π΅ and four πΆ, both of which are independent of π₯. So our constant term is negative three π΅ plus four πΆ.

To find the value of π΅, we rearrange this equation. And in doing so, we find that π΅ is equal to four πΆ plus two all over three. Remember, weβve already found the value πΆ. Itβs seven over 13. So substituting, we have that π΅ is equal to four multiplied by seven over 13 plus two all over three. Four multiplied by seven over 13 is 28 over 13. And then, we can think of the integer two as 26 over 13. We therefore have 28 over 13 plus 26 over 13 all over three, which is equivalent to 28 plus 26 over 13 multiplied by three. That gives 54 over 39 which simplifies to 18 over 13 by dividing both the numerator and denominator by three.

So weβve found the values of all three constants π΄, π΅, and πΆ, both by choosing values of π₯ which will eliminate certain parts of our expression and by comparing the coefficients of π₯ squared and the constant term. It would be a useful check of our work to make sure that the values weβve found for π΄, π΅, and πΆ do indeed work when comparing the final terms. So that would be the π₯ terms. Weβve found that the value of π΄ is negative seven over 13. The value of π΅ is 18 over 13. And the value of πΆ is seven over 13.