# How do you factor the expression x^2+3x+1?

Sep 2, 2016

${x}^{2} + 3 x + 1 = \left(x + \frac{3}{2} - \frac{\sqrt{5}}{2}\right) \left(x + \frac{3}{2} + \frac{\sqrt{5}}{2}\right)$

#### Explanation:

We can complete the square and use the difference of squares identity, which may be written:

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

with $a = \left(x + \frac{3}{2}\right)$ and $b = \frac{\sqrt{5}}{2}$ as follows:

${x}^{2} + 3 x + 1 = {\left(x + \frac{3}{2}\right)}^{2} - \frac{9}{4} + 1$

$\textcolor{w h i t e}{{x}^{2} + 3 x + 1} = {\left(x + \frac{3}{2}\right)}^{2} - {\left(\frac{\sqrt{5}}{2}\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} + 3 x + 1} = \left(\left(x + \frac{3}{2}\right) - \frac{\sqrt{5}}{2}\right) \left(\left(x + \frac{3}{2}\right) + \frac{\sqrt{5}}{2}\right)$

$\textcolor{w h i t e}{{x}^{2} + 3 x + 1} = \left(x + \frac{3}{2} - \frac{\sqrt{5}}{2}\right) \left(x + \frac{3}{2} + \frac{\sqrt{5}}{2}\right)$