3r^2+20r+32=0, solve for r

Question

Solve for r of the equation

Answer 1

3r²+20r+32=0.

Factors of 3=(1,3): factors of 32=(1,32), (2,16), (4,8).

(3r+a)(r+b)=3r²+r(a+3b)+ab, so (a,b)=(1,32), (2,16), (4,8), (32,1), (16,2) or (8,4), and:

a+3b=20:

a  b     3b  a+3b

1   32  96  97

2   16  48  50

4   8    24  28

8   4    12  20 is the correct combination

16 2      6  22

32 1      3  35

(3r+8)(r+4)=0 so 3r+8=0, 3r=-8, r=-8/3; or r+4=0, r=-4 (two solutions).