How do you factor completely #3x^2 - 5x - 2#?

1 Answer
Alan P.
Jun 23, 2016

#3x^2-5x-2=color(blue)((1x-2)(3x+1))#

Explanation:

Looking for #{a,b,c,d}# for a factoring #(ax+c)(bx+d)=3x^2-5x-2#

We see that
#color(white)("XXX")ab=3#
#color(white)("XXX")cd=-2#
and
#color(white)("XXX")ad+cb=-5#

In the hopes of finding integer values for #{a,b,c,d}#
we note that the factors of #3# are #{1,3}#
so if #a, b# are integers #{a,b}={1,3}#
and
we also note the the factors of #-2# are #{(-1,2),(1,-2)}#

We can test these possible combinations looking for #ad+cb=5#

#{: (underline(a),underline(b),underline(c),underline(d),,underline(ad+cb)), (1,3,-1,2,,-1), (1,3,2,-1,,+5), (1,3,1,-2,,+1), (color(red)(1),color(red)(3),color(red)(-2),color(red)(1),,color(red)(-5)) :}#

We have found values that match our requirements:
#color(white)("XXX")< a,b,c,d > = < 1,3,-2,1 >#
So
#color(white)("XXX")(ax+c)(bx+d)=(1x-2)(3x+1)#

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