# How do you factor -2x^2 + 1x + 1?

Nov 7, 2015
• Set the expression = 0 and use the ABC formula
• Find factors by using $\left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$
$- 2 {x}^{2} + x + 1 = \left(x + \frac{1}{2}\right) \left(x - 1\right)$

#### Explanation:

Start by setting the expression equal to zero, then use the ABC-formula.

The ABC-formula is:
$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Insert the values for a, b and c (which are the coefficients(red) in your expression: $\textcolor{red}{- 2} {x}^{2} + \textcolor{red}{1} x + \textcolor{red}{1}$

$x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \cdot \left(- 2\right) \cdot 1}}{2 \cdot \left(- 2\right)} \to \frac{- 1 \pm \sqrt{1 + 8}}{- 4} \to \frac{- 1 \pm 3}{-} 4$

${x}_{1} = \frac{- 1 + 3}{-} 4 = \frac{2}{-} 4 = - \frac{1}{2}$

${x}_{2} = \frac{- 1 - 3}{-} 4 = \frac{- 4}{- 4} = 1$

Put these values into $\left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

$\left(x - \left(- \frac{1}{2}\right)\right) \left(x - 1\right)$
$\left(x + \frac{1}{2}\right) \left(x - 1\right)$