# How do you multiply (sqrt5- sqrt2)(sqrt5+sqrt2)?

Jun 16, 2015

$\left(\sqrt{5} - \sqrt{2}\right) \left(\sqrt{5} + \sqrt{2}\right)$ is of the form

$\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$

So $\left(\sqrt{5} - \sqrt{2}\right) \left(\sqrt{5} + \sqrt{2}\right) = 5 - 2 = 3$

#### Explanation:

$\left(\sqrt{5} - \sqrt{2}\right) \left(\sqrt{5} + \sqrt{2}\right)$

$= {\left(\sqrt{5}\right)}^{2} - {\left(\sqrt{2}\right)}^{2} = 5 - 2 = 3$

using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Alternatively use FOIL to pick pairs of terms to multiply and add together:

First: $\sqrt{5} \cdot \sqrt{5} = 5$
Outside: $\sqrt{5} \cdot \sqrt{2}$
Inside: $- \sqrt{2} \cdot \sqrt{5}$ = $- \sqrt{5} \cdot \sqrt{2}$
Last: $- \sqrt{2} \cdot \sqrt{2} = - 2$

F + O + I + L = $5 + \cancel{\sqrt{5} \sqrt{2}} - \cancel{\sqrt{5} \sqrt{2}} - 2 = 3$

Jun 16, 2015

The answer is $3$.

#### Explanation:

You can use "FOIL" (First-Outside-Inside-Last) as follows:

$\left(\sqrt{5} - \sqrt{2}\right) \cdot \left(\sqrt{5} + \sqrt{2}\right) = \sqrt{5} \cdot \sqrt{5} + \sqrt{5} \sqrt{2} - \sqrt{5} \sqrt{2} - \sqrt{2} \sqrt{2}$

After cancellation and use of the facts that ${\sqrt{5}}^{2} = 5$ and ${\sqrt{2}}^{2} = 2$, this becomes $5 - 2 = 3$.

The reason "FOIL" works is because of abstract properties of numbers systems, such as the Distributive Property, which says that $a \cdot \left(b + c\right) = a \cdot b + a \cdot c$ for all numbers $a , b$, and $c$. You can use it twice when you expand $\left(a + b\right) \cdot \left(c + d\right)$ as follows:

$\left(a + b\right) \cdot \left(c + d\right) = \left(a + b\right) \cdot c + \left(a + b\right) \cdot d$

$= a \cdot c + b \cdot c + a \cdot d + b \cdot d$

This can be rearranged, by the Commutative Property, to $= a \cdot c + a \cdot d + b \cdot c + b \cdot d$. (FOIL)

It's also worthwhile to memorize the difference of two squares factoring formula: ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$. This can be applied to the problem at hand "in reverse" by letting $a = \sqrt{5}$ and $b = \sqrt{2}$.