### Video Transcript

The expression two π₯ plus one over π₯ plus two times π₯ plus three can be written in the form π΄ over π₯ plus three plus π΅ over π₯ plus two. Find the values of π΄ and π΅.

What weβre doing here is splitting this single fraction with two distinct linear factors in our denominator into two separate fractions with linear denominators. This is called splitting into partial fractions. In this case, weβre saying that two π₯ plus one over π₯ plus two times π₯ plus three is equal to π΄ over π₯ plus three plus π΅ over π₯ plus two for some constant values π΄ and π΅. So, how do we find the values of π΄ and π΅? Well, the first thing we want to do is kind of reverse this process. We want to make the expression on the right-hand side look a little more like that on the left.

And so, we recall that to add algebraic fractions or any fractions, in fact, we create a common denominator. Well, here the common denominator is π₯ plus three times π₯ plus two. And so to achieve this, we multiply the numerator and denominator of our first fraction by π₯ plus two and of our second by π₯ plus three. When we do, we get π΄ times π₯ plus two over π₯ plus two times π₯ plus three plus π΅ times π₯ plus three over π₯ plus two times π₯ plus three. Now, since the denominators are equal, we can simply add the numerators. So, we get π΄ times π₯ plus two plus π΅ times π₯ plus three over π₯ plus two times π₯ plus three.

Now, this is still equal to our earlier fraction. And if we look carefully, we should see that the denominators of our two fractions are equal. What this means is that the numerators themselves must also be equal. That is, two π₯ plus one must be equal to π΄ times π₯ plus two plus π΅ times π₯ plus three. And now we have two ways to work out the value of π΄ and π΅. One method is to distribute each of our parentheses and equate coefficients. That is, look at the coefficient of π₯ on the left- and right-hand side of our equation and look at the constants on the left- and right-hand side. What that will do is create a pair of simultaneous equations which we can solve for π΄ and π΅.

However, there is an easier way, that is, to substitute the zeros of π₯ plus two times π₯ plus three. Those are π₯ equals negative two and π₯ equals negative three. Now, when we do this, we end up creating an equation purely in terms of π΄ or π΅. Letβs see what that looks like. When we let π₯ be equal to negative two, the left-hand side becomes two times negative two plus one. Then, the right becomes π΄ times negative two plus two plus π΅ times negative two plus three. But of course, negative two plus two is zero. So, this becomes π΄ times zero, which is zero. And now we should see we have an equation purely in terms of π΅. Well, this simplifies to negative three equals π΅, and we therefore found the value of π΅.

Letβs repeat this process for π₯ equals negative three. The left-hand side of our equation becomes two times negative three plus one. And the right becomes π΄ times negative three plus two plus π΅ times negative three plus three. This time negative three plus three is zero. So, this term becomes π΅ times zero, which is zero. And our equation therefore simplifies to negative five equals negative π΄. By either multiplying or dividing through by negative one, we find π΄ to be equal to five. And so, we found the values of π΄ and π΅. π΄ is five and π΅ is negative three.

Now, in fact, usually we would look to express two π₯ plus one over π₯ plus two times π₯ plus three in partial-fraction form. And so, our next step might be to replace π΄ equals five and π΅ equals negative three in our earlier equation. And so, we find it can be written as five over π₯ plus three minus three over π₯ plus two.