### Video Transcript

Find π΄ and π΅ such that four over π₯ plus eight multiplied by π₯ minus two is equal to π΄ over π₯ minus two plus π΅ over π₯ plus eight.

What weβve been asked to find here is the partial fractions decomposition of the quotient on the left-hand side. Weβre separating out into the sum of two fractions, which have linear denominators. In order to find the values of π΄ and π΅, we first want to write each of the fractions on the right-hand side with the same denominator as we have on the left. So to create a common denominator of π₯ plus eight multiplied by π₯ minus two, we need to multiply both the numerator and denominator of the first fraction by π₯ plus eight and the numerator and denominator of the second fraction by π₯ minus two. Thatβs whichever linear factor isnβt already included in the denominator.

As the two fractions on the right-hand side will now have the same denominator, we can add them together. So the right-hand side becomes π΄ multiplied by π₯ plus eight plus π΅ multiplied by π₯ minus two all over π₯ plus eight multiplied by π₯ minus two. Now, as the fractions on each side of this equation are equal to one another and they now have the same denominator, we can write a new equation equating the numerators only. We have that four is equal to π΄ multiplied by π₯ plus eight plus π΅ multiplied by π₯ minus two. And We can use this equation to find the values of π΄ and π΅.

Remember, this equation is true for all values of π₯. So we can choose values of π₯ which will eliminate either π΄ or π΅ in turn in order to find the value of the other. Firstly, we can choose π₯ to be equal to negative eight because this gives the equation four equals π΄ multiplied by negative eight plus eight plus π΅ multiplied by negative eight minus two. And of course, negative eight plus eight is simply zero. So weβve eliminated the term in π΄. Weβre left with the equation four equals negative 10π΅ which we can solve by dividing both sides by negative 10. π΅ will be equal to negative four-10ths which simplifies to negative two-fifths.

The second value of π₯ we should chooses is the value that will make the second set of parentheses equal to zero. So we choose π₯ equals two, giving four equals π΄ multiplied by two plus eight plus π΅ multiplied by two minus two. But two minus two is intentionally equal to zero. We have, therefore, eliminated the π΅s and we have the equation four is equal to 10π΄. We solved by dividing both sides by 10, giving π΄ equals four-10ths which simplifies to two-fifths. So by considering the partial fractions decomposition of four over π₯ plus eight multiplied by π₯ minus two, weβve found that the value of π΄ is two-fifths and the value of π΅ is negative two-fifths.