# Week 15 – precalculus 11

Multiplying and Dividing Rational Expressions:

Multiplying rational expressions is just like multiplying any other fraction, just multiply the top and bottom. Although if we are part of the expression is in a different form, we will need to factor it before multiplying. When it comes to division we just recipricate the fraction or “flip” it and multiply, we must remember the value of x on the bottom if there is one before recipricating so the bottom never equals 0. These values are called non-permissible values and it represents values that x cannot be to prevent there being 0 on the bottom of a fraction, which is not allowed. We also do this for any other variables on the denominator. We basically find what would make a set equal 0. If x is a coefficient it is automatically a non-permissible value of 0. To demonstrate further I will simplify 3 expressions, one simple muliplication, one more complex muliplication, and one division. To do all of these we must remember 3 steps: Factor, find the non permissible values, and then cancel out values.

We will start by simplifying $\frac {p^2}{10}\times \frac {20}{p}$

Because we cannot factor and there are not common factors to cancel out, we will multiply, resulting in:

$\frac {20p^2}{10p}$

The non permissible values so far are just 0

We can simplify this further but dividing by p and 10, giving us

$2p$

And to state what the non-permissible values are we say: $x \neq 0$

Now we will do a more complex multiplication expression:

$\frac {(x-3)(x+2)}{x+4}\times \frac {x^2 - 16}{x(x+2)}$

Now in this case we only need to factor one value but sometimes we will need to factor all of them.

$\frac {(x-3)(x+2)}{x+4}\times \frac {(x+4)(x-4)}{x(x+2)}$

We will now make a giant fraction, we can do this because if they are beside eachother they are multiplying anyway.

$\frac {(x-3)(x+2)(x+4)(x-4)}{(x+4) x(x+2)}$

This is also where we should find the non-premissible values, which are 0, -4, -2

Now when there is a plus or minus sign the to values it applies to are now connected and become basically one value. In this case we can cancel out x+2 and x+4 because they can divide into one, we still include them in the non-permissible values however.

We also do not foil the top values, it is simplified.

$\frac {(x-3)(x-4)}{x}$, $x \neq 0,-4,-2$

We will now do one last expression, it will be a more simple division expression just so you get the idea. The division expression gets turned into a multiplication so we only need to add an extra step.

$\frac {10}{x^2} \div \frac {5}{x^2}$

Now in this case, x cannot equal 0, we must take a non-permissible value from x if it is on the bottom before we flip the expression, we also take non-permissible values of things we have cancelled.

Now all we do its just flip the second fraction

$\frac {10}{x^2} \times \frac {x^2}{5}$

The non-permissible value is still just 0

Now we will muliply, this is a more simple expression so we can just muliply because we cannot factor like a quadratic.

$\frac {10x^2}{5x^2}$

Now we can also divide this to a certain extent to further simplify, we will divide by $5x^2$ giving us

$2$, $\neq 0$

And that is how you multiply and divide rational expressions