Solve by factoring? #12x^2-37x-780 =0#

1 Answer
Sep 1, 2017

#(3x+20)(4x-39)=0#

#x =-20/3" or " x = 39/4#

Explanation:

First determine whether there are rational factors.

#a = 12" "b = -37 " "c = -780#

#b^2 - 4ac = (-37)^2 -4(12)(-780) = 38809 = 197^2#

We need to find factors of #12 and 780# whose products differ by #37#

The smaller the value if #b#, the closer the factors are to #sqrt(ac)#

#sqrt(12 xx 780) = sqrt9360 = 96.7 ~~97#

The factors will be on either side of #96.7#, about #37/2# bigger or smaller.

#37/2 = 18.5#

#97 -18 =79 and 97+18 = 115#

Use trial and error with values close to #79 and 115#

#9360# is not divisible by #79#

#9360 div 80 = 117" "# BINGO! #" "117-80 = 37#

These are the factors, now we need to form them from #12 and 780#

#" "12 and 780#
#" "darrcolor(white)(xxx)darr#
#" "3color(white)(xxxx)20" "rarr 4 xx20 = 80#
#" "4color(white)(xxxx)39" "rarr 3xx39 = ul117#
#color(white)(xxxxxxxxxxxxxxxxxxxxx)37#

These are the factors, now fill in the signs to get #-37 and -780#

#" "12 and 780#
#" "darrcolor(white)(xxx)darr#
#" "3color(white)(xxxx)+20" "rarr 4 xx+20 = +80#
#" "4color(white)(xxxx)-39" "rarr 3xx-39 = -ul117#
#color(white)(xxxx.xxxxxxxxxxxxxxxxxxxx)-37#

#(3x+20)(4x-39)=0#

Set each factor equal to #0#

#3x+20 =0 " "rarr x =-20/3#

#4x-39 =0" "rarr x = 39/4#