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

1 Answer
Alan P.
Mar 27, 2015

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#

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