Simplifying Radical Expressions Involving Products

When you're multiplying terms that contain radicals, it's important to remember when the properties of square roots can be applied to combine terms, and when terms need to be kept separate.

Example 1:

Simplify.

$(3 + \sqrt{2}) (11 + \sqrt{5})$

Multiply using FOIL (First, Outer, Inner, Last).

$\begin{array}{l} = 3 \cdot 11 + 3 \cdot \sqrt{5} + \sqrt{2} \cdot 11 + \sqrt{2} \cdot \sqrt{5} \\ = 33 + 3 \sqrt{5} + 11 \sqrt{2} + \sqrt{10} \end{array}$

In the last expression above, all of the terms have different radicals, and none of the radicands has any perfect square factors. So the expression cannot be simplified further.

Since a negative number times a negative number is always a positive number, you need to remember when taking a square root that the answer will be both a positive and a negative number or expression. For example $a \cdot a = a^{2}$ , and also $(- a) \cdot (- a) = a^{2}$ .We usually will denote such dual answers as $\pm a$ .

Example 2:

Simplify.

$\sqrt{6} (1 + 5 \sqrt{3})$

$\begin{array}{l} = \sqrt{6} + 5 \cdot \sqrt{6 \cdot 3} \\ = \sqrt{6} + 5 \sqrt{18} \end{array}$

The radicand $18$ here contains a perfect square factor, so the expression can be simplified further.

$\begin{array}{l} = \sqrt{6} + 5 \sqrt{9 \cdot 2} \\ = \sqrt{6} + 5 \sqrt{9} \cdot \sqrt{2} \\ = \sqrt{6} + 5 (\pm 3) \cdot \sqrt{2} \\ = \sqrt{6} \pm 15 \sqrt{2} \end{array}$

Often, you may need to square a binomial involving one or more radical terms.

Example 3:

Simplify.

${(5 - \sqrt{6})}^{2}$

$= (5 - \sqrt{6}) (5 - \sqrt{6})$

Multiply using FOIL (First, Outer, Inner, Last).

$= 5 \cdot 5 + 5 \cdot (- \sqrt{6}) + (- \sqrt{6}) \cdot 5 + (- \sqrt{6}) \cdot (- \sqrt{6})$

Combine like terms.

$\begin{array}{l} = 25 - 5 \sqrt{6} - 5 \sqrt{6} + 6 \\ = 31 - 10 \sqrt{6} \end{array}$

Example 4:

Simplify.

$\sqrt{5} {(3 + \sqrt{10})}^{2}$

$= \sqrt{5} (3 + \sqrt{10}) (3 + \sqrt{10})$

Multiply using FOIL (First, Outer, Inner, Last).

$\begin{array}{l} = \sqrt{5} (3 \cdot 3 + 3 \cdot \sqrt{10} + \sqrt{10} \cdot 3 + \sqrt{10} \cdot \sqrt{10}) \\ = \sqrt{5} (9 + 3 \sqrt{10} + 3 \sqrt{10} + \sqrt{100}) \\ = \sqrt{5} (9 + 6 \sqrt{10} + 10) \\ = \sqrt{5} (19 + 6 \sqrt{10}) \\ = 19 \sqrt{5} + 6 \sqrt{10} \cdot \sqrt{5} \\ = 19 \sqrt{5} + 6 \sqrt{50} \end{array}$

Note that $50$ has a perfect square factor.

$\begin{array}{l} = 19 \sqrt{5} + 6 \sqrt{25} \cdot \sqrt{2} \\ = 19 \sqrt{5} + 6 (\pm 5) \cdot \sqrt{2} \\ = 19 \sqrt{5} \pm 30 \sqrt{2} \end{array}$

Simplifying Radical Expressions Involving Products

Download our free learning tools apps and test prep books