# How to factor 2x^2+3x+1=0 ?

Dec 11, 2017

$\left(2 x + 1\right) \left(x + 1\right) = 0$
$\implies x = - \frac{1}{2} , x = - 1$

#### Explanation:

To solve this we must understand how to factor:

if $y = a {x}^{2} + b x + c$

Then you need to find two numbers who sum to make $b$ but multiply to make $a \cdot c$

So in this circumstance:

Plus to make $3$ and multiply to make $2 \cdot 1 = 2$

Pairs of numbers who multiply to make $2$:

$\left(2 , 1\right) , \left(- 2 , - 1\right)$

But out of these we see that $\left(2 , 1\right)$ also plus to make $3$

So we can use the fact that $2 x + x = 3 x$ :

$\implies 2 {x}^{2} + 2 x + x + 1$

Now we can factor the first 2 and last 2 terms individually, but ensuring they both have the same factor:

$\implies 2 x \left(x + 1\right) + \left(x + 1\right)$

Factoring out the $\left(x + 1\right)$ :

$\implies \left(x + 1\right) \left[2 x + 1\right]$

$\implies \left(x + 1\right) \left(2 x + 1\right)$

If we want to solve for $0$ then either one of $\left(x + 1\right) \mathmr{and} \left(2 x + 1\right)$ must equal $0$ and $\alpha \cdot 0 = 0$

So hence $\left(x + 1\right) = 0 \implies x = - 1$

Also $\left(2 x + 1\right) = 0 \implies 2 x = - 1 \implies x = - \frac{1}{2}$