3r^2 - 16r - 7 = 5 subtract 5 from both sides
3r^2 - 16r - 12 = 0 we cannot take out any common factor
So....here is a procedure for factoring whenever the lead coefficient is not a 1
1) Multiply the first and last terms = 3 * -12 = -36
2) List all the possible factors of -36 and that add to - 16.....we can stop (if) we find one that adds to the middle term, -16......if we find no combination.....the polynomial is fully factored
-36 1
-18 2
Note that -18 and 2 is what we need
3) Write the original problem in two ways
3x^2 - 18x + 2x - 12 = 0 or
3x^2 + 2x - 18x - 12 = 0
The second can't be factored by grouping, but the first one can
So we have
3x^2 - 18x + 2x - 12 = 0 factor by grouping
3x ( x - 6) + 2 ( x - 6) = 0 the GCF is (x - 6)
( x - 6) ( 3x + 2) = 0 so
x - 6 = 0 or 3x + 2 = 0
x = 6 subtract 2 from both sides
3x = - 2
divide both sides by 3
x = -2/3
The solutions are in red