Question

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

Answer 1

`(2x+1)(x+1) = 0`
`=> x = -1/2 , x= -1`

Explanation:

To solve this we must understand how to factor:

if `y = ax^2 + bx + c`

Then you need to find two numbers who sum to make `b` but multiply to make `a*c`

So in this circumstance:

Plus to make `3` and multiply to make `2*1 = 2`

Pairs of numbers who multiply to make `2`:

`( 2 , 1) , (-2 ,-1 )`

But out of these we see that `(2,1)` also plus to make `3`

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

`=> 2x^2 + 2x + x +1`

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

`=> 2x(x+1 ) + (x+1)`

Factoring out the `(x+1)` :

`=> (x+1)[2x + 1 ]`

`=> (x+1)(2x+1)`

If we want to solve for `0` then either one of `(x+1) or (2x+1)` must equal `0` and `alpha*0 = 0`

So hence `(x+1) = 0 => x = -1`

Also `(2x+1) = 0 => 2x = -1 => x = -1/2`