# How do you factor 6x^2+23x+6?

Jun 4, 2016

Root1 $= - 0.282 \left(3 \mathrm{dp}\right)$ & Root2 $= - 3.552 \left(3 \mathrm{dp}\right)$

#### Explanation:

This is a quadratic equation of form $a {x}^{2} + b x + c$ Here a=6;b=23;c=6 Roots are $- \frac{b}{2 a} \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a} \mathmr{and} - \frac{23}{12} \pm \frac{\sqrt{529 - 144}}{12} = - 1.916 \pm \frac{19.62}{12} = - 1.916 \pm 1.635 \therefore$ Root1 $= - 0.282$ & Root2 $= - 3.552$[Ans]

Jun 4, 2016

$6 {x}^{2} + 23 x + 6 = 6 \left(x + \frac{23 - \sqrt{385}}{12}\right) \left(x + \frac{23 + \sqrt{385}}{12}\right)$

#### Explanation:

Complete the square, then use the difference of squares identity:

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

with $a = \left(x + \frac{23}{12}\right)$ and $b = \frac{\sqrt{385}}{12}$ as follows:

$6 {x}^{2} + 23 x + 6$

$= 6 \left({x}^{2} + \frac{23}{6} x + 1\right)$

$= 6 \left({\left(x + \frac{23}{12}\right)}^{2} - {23}^{2} / {12}^{2} + 1\right)$

$= 6 \left({\left(x + \frac{23}{12}\right)}^{2} - \frac{529 - 144}{12} ^ 2\right)$

$= 6 \left({\left(x + \frac{23}{12}\right)}^{2} - \frac{385}{12} ^ 2\right)$

$= 6 \left(\left(x + \frac{23}{12}\right) - \frac{\sqrt{385}}{12}\right) \left(\left(x + \frac{23}{12}\right) + \frac{\sqrt{385}}{12}\right)$

$= 6 \left(x + \frac{23 - \sqrt{385}}{12}\right) \left(x + \frac{23 + \sqrt{385}}{12}\right)$

Note that it is not possible to simplify $\sqrt{385}$ any further since $385 = 5 \cdot 7 \cdot 11$ has no square factors.