### 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 ππ₯ 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. Letβs 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.

Weβll 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.