# How do you write y=5x^2+5x-3 in factored form?

Sep 22, 2015

You'll need to find the zeros first.

#### Explanation:

We cannot factor $y = 5 {x}^{2} + 5 x - 3$ by trial and error, or by AC, or by grouping.

Solve: $5 {x}^{2} + 5 x - 3 = 0$ by either completing the square or by using the quadratic formula to get solutions

${x}_{1} = \frac{- 5 + \sqrt{85}}{10}$ and ${x}_{2} = \frac{- 5 - \sqrt{85}}{10}$

The factors of $5 {x}^{2} + 5 x - 3$ are $x - {x}_{1}$ and $x - {x}_{2}$, so we have the factored form:

$y = \left(x - \frac{- 5 + \sqrt{85}}{10}\right) \left(x - \frac{- 5 - \sqrt{85}}{10}\right)$

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If you wanted vertex form, here is the answer

$y = 5 \left({x}^{2} + x \textcolor{w h i t e}{\text{xxxxxx}}\right) - 3$

$y = 5 \left({x}^{2} + x + \frac{1}{4}\right) - 3 - \frac{5}{4}$

$y = 5 {\left(x - \frac{1}{2}\right)}^{2} - \frac{17}{4}$

Refer to explanation

#### Explanation:

First we have to find the roots of $5 {x}^{2} + 5 x - 3 = 0$. Hence the roots are

${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

where $a = 5 , b = 5 , c = - 3$ so we have that

${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} \implies {x}_{1 , 2} = \frac{- 5 \pm \sqrt{{5}^{2} + 4 \cdot 5 \cdot 3}}{10} = \frac{- 5 \pm \sqrt{85}}{10}$

Hence the factor form is

y=a*(x-x_1)*(x-x_2)=5*(x-(-5+sqrt85)/10)*(x-(-5-sqrt85)/10)= 1/20*(10x+5-sqrt85)*(10x+5+sqrt85)