# How do you factor 2x^2+x-5?

Jun 22, 2017

$2 {x}^{2} + x - 5 = \textcolor{red}{2 \left(x + \frac{1 + \sqrt{41}}{4}\right) \left(x + \frac{1 - \sqrt{41}}{4}\right)}$

#### Explanation:

$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$\textcolor{w h i t e}{\text{XXXXXXXXX}}$ where $a {x}^{2} + b x + c = 0$

to $2 {x}^{2} + x - 5 = 0$

obtaining the solutions
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- 1 \pm \sqrt{41}}{4}$
and thus the factors
$\textcolor{w h i t e}{\text{XXX}} \left(x - \frac{- 1 + \sqrt{41}}{4}\right) \mathmr{and} \left(x - \frac{- 1 - \sqrt{41}}{4}\right)$

Since
$\textcolor{w h i t e}{\text{XXX}} \left(x + \frac{1 - \sqrt{41}}{4}\right) \left(x + \frac{1 + \sqrt{41}}{4}\right) = {x}^{2} + \frac{x}{2} - \frac{5}{2}$
we will have an additional constant factor of $2$

Jun 22, 2017

$f \left(x\right) = 2 \left(x + \frac{1}{4} - \frac{\sqrt{41}}{4}\right) \left(x + \frac{1}{4} + \frac{\sqrt{41}}{4}\right)$

#### Explanation:

$f \left(x\right) = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) .$

${x}_{1} \mathmr{and} {x}_{2}$ are the $2$ real roots of $f \left(x\right) = 0$.

They are given by the improved quadratic formula (Socratic Search):

$x = - \frac{b}{2 a} \pm \frac{d}{2 a}$

In this case:

$D = {b}^{2} - 4 a c = 1 + 40 = 41 \rightarrow d = \pm \sqrt{41}$
$x = - \frac{1}{4} \pm \frac{\sqrt{41}}{4}$
${x}_{1} = - \frac{1}{4} + \frac{\sqrt{41}}{4}$
${x}_{2} = - \frac{1}{4} - \frac{\sqrt{41}}{4}$

$f \left(x\right) = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

$f \left(x\right) = 2 \left(x + \frac{1}{4} - \frac{\sqrt{41}}{4}\right) \left(x + \frac{1}{4} + \frac{\sqrt{41}}{4}\right)$