Question

How do you factor `2x^2 + 13x + 6 = 0`?

Answer 1

Assuming `2x^2 + 13x + 6` can be factored into
`(ax+b)(cx+d)` with all integer values for `a, b, c, d`

there are a very limited number of possibilities
(for example `{a,c} = {1,2}` )
and it should be possible to factor this by simple trial and error as
`(2x+1)(x+6) = 0`

If the quadratic were not so simple you could use the observation
that the quadratic equation is of the form
`ax^2+bx+c = 0`
and use the formula
`x = (-b+-sqrt(b^2-4ac))/(2a)`
to give the values for which the quadratic equals `0`.

Supposing that this method gave values
`x=c` and `x=d`
then the factoring of given equation would be
`(k)(x-c)(x-d) = 0`
where `k` is a constant (in this case `= 2`) needed to convert the factored form into the original quadratic expression.
For example, in this specific case (without showing the details) we would get:
`x=-6` and `x-1/2`
and
`(x+6)(x+1/2) = x^2 + 6 1/2x +3`
which needs to be multiplied by `2`
to get the quadratic in its original form"
`2x^2+13x+6`