# Week 10 – precalculus 11

Dividing radicals is not as easy as multiplying, though it does include multiplying. We will need to use distributive property and potentially adding to rationlize the denominator. We cannot divide by a radical so we must make the denominator a whole number. We normally multiply the denominator by itself and multiply that with the top as well but if we have a binomial we must multiply the top and bottom by the conjugate of the denominator. A conjugate is the same as the denominator but the plus or minus sign will be the opposite of the denominator. To see how this works we must show it. The goal is to simplify the expression as much as possible. We already know how to multiply radicals, this is just combining those skills with some adding.

First we will start with an expression like $\frac {4\sqrt {2} - 6\sqrt {5}}{2\sqrt {3}}$

Because it is a monomial denominator we will multiply the top and bottom by $2\sqrt{3}$

$\frac {8\sqrt {6} - 12\sqrt {15}}{4\sqrt {9}}$

The denominator now should become 3×4

$\frac {8\sqrt {6} - 12\sqrt {15}} {12}$

Because we cannot simply the numerators values or add them together we will simply divide them both by a common factor as with the denominator. Giving us:

$\frac {2\sqrt {6} - 3\sqrt {15}} {3}$

Let’s say that we are given a expression with a binomial denominator.

We are given $\frac {5\sqrt {3} + 4\sqrt {2}}{2\sqrt {3} - \sqrt 2}$

To solve this we will have to multply the top and bottom by the conjugate of the denominator.

We are given $\frac {(5\sqrt {3} + 4\sqrt {2}) (2\sqrt {3}+ \sqrt {2})}{(2\sqrt {3} - \sqrt 2) (2\sqrt {3}+ \sqrt {2})}$

Then we will foil/muliply

$\frac {10\sqrt {9} + 5\sqrt {6} + 8\sqrt {6} + 4\sqrt {4}}{4\sqrt {9} + 2\sqrt {6} - 2\sqrt {6} - \sqrt {4}}$

Many of these we can either add together or turn into whole numbers. We will also add together the whole numbers. Two of the radicals on the denominator cancel out.

$\frac {38 + 13\sqrt {6}}{10}$

At this point, the numbers cannot be divided by a common number. This is the simplified form. This it how you use multplication to rationalize a denominator and simplify a radical expression in a division setting.