# How do you write y=2x^2-10x+5 in factored form?

Sep 20, 2015

$y = 2 \left(x - \frac{5 + \sqrt{15}}{2}\right) \left(x - \frac{5 - \sqrt{15}}{2}\right)$

#### Explanation:

$2 {x}^{2} - 10 x + 5 = 2 \left({x}^{2} - 5 x + \frac{5}{2}\right) =$

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

$2 \left({\left(x - \frac{5}{2}\right)}^{2} - \frac{25}{4} + \frac{10}{4}\right) = 2 \left({\left(x - \frac{5}{2}\right)}^{2} - \frac{15}{4}\right) =$

$= 2 \left({\left(x - \frac{5}{2}\right)}^{2} - {\left(\frac{\sqrt{15}}{2}\right)}^{2}\right) =$

$= 2 \left(x - \frac{5}{2} - \frac{\sqrt{15}}{2}\right) \left(x - \frac{5}{2} + \frac{\sqrt{15}}{2}\right) =$

$= 2 \left(x - \frac{5 + \sqrt{15}}{2}\right) \left(x - \frac{5 - \sqrt{15}}{2}\right)$

General case:

${P}_{2} \left(x\right) = a {x}^{2} + b x + c = a \left({x}^{2} + \frac{b}{a} x + \frac{c}{a}\right) =$

$= a \left({x}^{2} + 2 \cdot x \cdot \frac{b}{2 a} + {\left(\frac{b}{2 a}\right)}^{2} - {\left(\frac{b}{2 a}\right)}^{2} + \frac{c}{a}\right) =$

$= a \left({\left(x + \frac{b}{2 a}\right)}^{2} - \frac{{b}^{2} - 4 a c}{4 {a}^{2}}\right) =$

$= a \left({\left(x + \frac{b}{2 a}\right)}^{2} - {\left(\frac{\sqrt{{b}^{2} - 4 a c}}{2 a}\right)}^{2}\right) =$

$= a \left(x + \frac{b}{2 a} - \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}\right) \left(x + \frac{b}{2 a} + \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}\right) =$

$= a \left(x - \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a}\right) \left(x - \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a}\right)$

You can see that ${P}_{2} \left(x\right) = 0$ for

$x - \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a} = 0 \implies x = \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a}$

or

$x + \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a} = 0 \implies x = \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a}$

which is famous formula for solutions of the quadratic equation.