How do you factor #3x^2 - 5x - 12#?

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Answer 1 · 2017-08-04T11:24:17

The answer is #(x-3)(3x+4)#.

![https://useruploads.socratic.org/uA9v8owQVa0rRLtH5qXG_1501845782432413515707.jpg)

Answer 2 · 2017-08-04T12:37:04

#3x^2−5x−12#

#=(x-3)(3x+4)#

Another Method
Using Quadratic Formula,
#(-b+-sqrt(b^2-4ac))/(2a)#

#a=3#
#b=-5#
#c=-12#

Applying Formula we get,

#(-(-5)+-sqrt(-5^2-4(3)(-12)))/(2(3)#
#(+5+-sqrt(25+144))/(6)#
#(5+-sqrt(169))/(6)#

#(5+-(13))/(6)#

#x=(5+13)/(6)=18/6=3# _ here #(x-3)#
and
#x=(5-13)/(6)=-8/6=-4/3# _ here #(3x+4)#

Answer 3 · 2017-08-04T13:07:48

#(x-3)(3x+4)#

#3x^2 -5x -12# is a quadratic trinomial.

Find factors of #3 and 12# whose products differ by #5#

#" "3 " & " 12#
#" "darr" "darr#

#" "1" "3" "rarr3 xx 3 =9 #
#" "3" "4" "rarr1xx4 = ul4#
#color(white)(w.wwwwwwwwww..wwww)5#

We have correct factors, now fill in the signs to get#-5 and 12#

#" "1" "-3" "rarr3 xx -3 =-9 #
#" "3" "+4" "rarr1xx+4 = +ul4#
#color(white)(w.wwwwwwwwwwwwww.wwww)-5#

#3x^2 -5x -12#

#=(x-3)(3x+4)#