# How do you factor 12p^2+16p+5?

Jun 2, 2015

$\left(6 p + 5\right) \left(2 p + 1\right)$

Use the quadratic formula for solving $a {x}^{2} + b x + c = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Taking ${b}^{2} - 4 a c$ this is ${16}^{2} - 4.12 .5 = 256 - 240 = 16$, $\sqrt{16} = 4$

This gives $p = \frac{- 16 \pm 4}{2.12}$ thus $p = - \frac{12}{24} \mathmr{and} - \frac{20}{24}$

Simplifying gives $p = - \frac{1}{2} \mathmr{and} p = - \frac{5}{6}$

These can be written as $2 p + 1 = 0 \mathmr{and} 6 p + 5 = 0$

Hence $12 {p}^{2} + 16 p + 5 = \left(6 p + 5\right) \left(2 p + 1\right)$