### Video Transcript

Express π₯ squared minus two over π₯ plus two times π₯ minus three times π₯ plus one in partial fractions.

A single fraction with two or more distinct linear factors in the denominator can be split into two or more separate fractions with linear denominators. Thisβs called splitting it into partial fractions. In this case, we have one, two, three distinct linear factors in the denominator of our fraction. And so, we can rewrite the entire expression as the sum of three partial fractions. We call these π΄ over π₯ plus two, π΅ over π₯ minus three, and πΆ over π₯ plus one, where π΄, π΅, and πΆ are real constants.

Then, our next job is to make the expression on the right-hand side look a little more like that on the left. And so, we recall that when we add algebraic fractions or any fraction, we create a common denominator. The common denominator is the product of our three denominators. Itβs π₯ plus two times π₯ minus three times π₯ plus one. And so, to achieve this common denominator, we multiply the numerator and denominator of our first fraction by π₯ minus three times π₯ plus one. We multiply the numerator and denominator of our second fraction by π₯ plus two times π₯ plus one. And of our third fraction by π₯ plus two times π₯ minus three.

And so, we get π΄ times π₯ minus three times π₯ plus one as one of our numerators. Our second numerator is π΅ times π₯ plus two times π₯ plus one. And our third is πΆ times π₯ plus two times π₯ minus three. All three of these are over π₯ plus two times π₯ minus three times π₯ plus one. And so, it follows that we can simplify a little by simply adding the numerators. Now, which is still, of course, equal to our earlier fraction.

And we can see, if we look carefully, that the denominators in each of these fractions are equal. Now, what this actually tells us is that for the two fractions to be equal, their numerators must themselves be equal. That is, π₯ squared minus two must be equal to π΄ times π₯ minus three times π₯ plus one plus π΅ times π₯ plus two times π₯ plus one plus πΆ times π₯ plus two times π₯ minus three.

Now, we have two different ways in which we can find the values of π΄, π΅, and πΆ. One method we have is to distribute our parentheses. And what we then do is we equate coefficients for π₯ squared, π₯, and the constants. That can be quite a long-winded process though. So, there is another method. The other method is to substitute the zeros of π₯ plus two times π₯ minus three times π₯ plus one into the equation. Those zeros are π₯ equals three, π₯ equals negative one, and π₯ equals negative two.

Now, the purpose of doing this is to create equations purely in terms of π΄, π΅, and πΆ, which we can then solve. So, weβll begin by letting π₯ be equal to three. The left-hand side of our equation becomes three squared minus two. Then, the right becomes π΄ times zero times four plus π΅ times five times four plus πΆ times five times zero. This simplifies to seven equals 20π΅. And dividing through by 20, we find π΅ is equal to seven twentieths. We can replace this in partial fraction form. Now, we wonβt leave it as seven twentieths over π₯ minus three when we answer this question.

But for now, letβs let π₯ be equal to negative one. The left-hand side of our equation becomes negative one squared minus two. And the right becomes π΄ times negative four times zero plus π΅ times one times zero plus πΆ times one times negative four. This simplifies to negative one equals negative four πΆ. And dividing through by negative four, and we find πΆ is equal to one-quarter.

Our last job is to let π₯ be equal to negative two. Weβre hoping here that weβll find the value of π΄. On the left-hand side, we get negative two squared minus two. And on the right, we get π΄ times negative five times negative one plus π΅ times zero times negative one plus πΆ times zero times negative five. This equation simplifies to two equals five π΄. And dividing through by five, we find π΄ is equal to two-fifths.

And so, we have two-fifths over π₯ plus two plus seven twentieths over π₯ minus three plus a quarter over π₯ plus one. Now, of course, we can rewrite this a little as two over five times π₯ plus two plus seven over 20 times π₯ minus three plus one over four times π₯ plus one. Itβs important to realise that these can be written in any order. So, to match the answer in this question, we get one over four times π₯ plus one plus two over five times π₯ plus two plus seven over 20 times π₯ minus three.