Question 28176
m^2 + m - 72 = 0----(1)
m^2 + [(9m) +(-8m)] - 72 = 0
[m^2+9m]-8m-72=0  (by additive associativity)
m(m+9)-8(m+9) = 0
mp -8p = 0 where p = (m+9)
p(m-8) = 0
(m+9)(m-8)=0
(m+9) = 0 gives m = -9
(m-8) = 0 gives m = 8
Answer: m = -9 and m = 8
Put m=-9 on the LHS of (1)and get 81-9-72 = 0 =RHS
m= 8 on the LHS of (1)and get 64+8-72 = 0 =RHS
Therefore our values   m = -9 and m = 8 are correct

NOTE:
[The golden rule is: write the mid term as the sum of two terms whose product is the product of the square term and the constant term]
[HOW? Multiply the square term and the constant term: (m^2)X(-72) = -72m^2
And (-72) = -(1X2X2X2X3X3) = (9)X(-8) so that the mid term: (m) =[(9m) +(-8m)] and the product of these two parts:(9m) and (-8m) = -72m^2  using the  concept
If the sign of the product is minus, and the  middle term if +,then give the sign of the midterm to the larger set and the other sign to the smaller.