### Video Transcript

Simplify π of π₯ equals three over π₯ plus two plus two π₯ over π₯ squared minus four.

In this question, weβre given a function π of π₯, which is the sum of two algebraic fractions or quotients. In order to simplify π of π₯, we need to try and combine these two fractions into a single fraction. We know that in order to add two fractions together, they must have a common denominator.

Letβs look at the denominators of these two fractions a little more closely. The first denominator is a linear function of π₯, itβs π₯ plus two, whereas the second denominator is a quadratic, itβs π₯ squared minus four. We should notice though that π₯ squared minus four is a particular type of quadratic. It is a difference of two squares, which we can abbreviate to DOTS. We know that if we have a difference of two squares, π₯ squared minus π squared, then this can be factored as π₯ minus π multiplied by π₯ plus π. As four is equal to two squared, we know that π₯ squared minus four can therefore be factored as π₯ minus two multiplied by π₯ plus two.

Now, we see that one of these two linear factors is in fact shared with the denominator of the first fraction. In order to see how we should proceed from here, letβs use a numeric example. Suppose we were adding the two fractions one-third and one-sixth. In order to do this, weβd need to convert both fractions to equivalent fractions with a common denominator. The denominator we would use would be the lowest common multiple of the two denominators themselves. And as three is a factor of six, that lowest common multiple would be six.

This gives us a clue for how we can proceed in our algebraic example. We need the lowest common multiple of π₯ plus two and π₯ minus two multiplied by π₯ plus two. But as one denominator is a factor of the other, the lowest common multiple of these will in fact just be the second denominator. Itβs π₯ minus two multiplied by π₯ plus two. So, in fact, we wonβt need to convert the second fraction at all, just as we didnβt need to convert the second fraction in our numeric example.

Letβs take the expression for π of π₯. In order to write the first fraction as an equivalent fraction with a denominator of π₯ plus two multiplied by π₯ minus two, we have to multiply both the numerator and denominator by π₯ minus two. As weβve already said though, in the second fraction, we donβt need to do anything, because this already has the denominator we want. As the two fractions now have the same denominator, we can add them by adding the numerators, giving three multiplied by π₯ minus two plus two π₯ all over π₯ plus two multiplied by π₯ minus two.

We can then distribute the parentheses in the numerator to give three π₯ minus six plus two π₯ over π₯ plus two multiplied by π₯ minus two and finally simplify by collecting the like terms in the numerator to give five π₯ minus six over π₯ plus two multiplied by π₯ minus two. This fraction canβt be simplified any further as there are no common factors in the numerator and denominator.

And so we have our answer for the simplified form of the function π of π₯. π of π₯ is equal to five π₯ minus six over π₯ plus two multiplied by π₯ minus two. And we can write those factors in the denominator in either order.