How do you factor $2 x^{2} + 13 x + 6 = 0$ $2x^2 + 13x + 6 = 0$ ?

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How do you factor $2 x^{2} + 13 x + 6 = 0$ $2x^2 + 13x + 6 = 0$ ?

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Answer 1 · 2015-03-27T15:43:19

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

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

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

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

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How do you factor $2 x^{2} + 13 x + 6 = 0$ $2x^2 + 13x + 6 = 0$ ?

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How do you factor $2 x^{2} + 13 x + 6 = 0$ $2x^2 + 13x + 6 = 0$ ?