How do you factor #16 = 38r - 12r^2#?

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Answer 1 · 2017-08-07T06:55:41

#2(2r-1)(3r-8)=0#

Set it equal to #0#
Take out the common factor of #2# first.

#12r^2-38r+16 =0#

#2(6r^2-19r+8)=0#

Find factors of #6 and 8# whose products ADD to #-19#

As a hint, we see that #19# is odd. Therefore the factors of #8# cannot be #2 and 4#, because their products will be even, so use #1 and 8#

#" "6 and 8#
#" "darr" "darr#
#" "2" "1" "rarr3xx1 = 3#
#" "3" "8" "rarr2xx8=ul16#
#color(white)(wwwwwwwwwwwwwwww)19#

The signs in the brackets will both be negative.

#2(2r-1)(3r-8)=0#

These are the factors which will lead to solutions of:

#r = 1/2 and r = 8/3#