Question 27742
Factor:
How do I do this ?

Solve:
(z-5)(z+4)=52
Thanks for your help and time, it's greatly appreciated !

Consider the given equation
(z-5)(z+4)=52  ----(1)
Expanding the product on the LHS, we have
(z-5)(z+4)=z(z+4)-5(z+4) = z^2 +4z -5z -(5x4) = z^2 +(4z-5z)-20 = z^2-z-20 --(*)
Using * in (1), we have
z^2-z-20 =52
z^2-z-20-52 = 0
z^2-z-72 = 0  ----(2)which is a quadratic in z
The product of the coefficient of the square term and the constant term is
(1)X(-72) = -72 = -(1X2X2X2X3X3)
The numerical factors are 1,2,2,2,3 and 3
We may leave out the 1 because 1X(anything)= the same thing
Grouping the factors into two sets(not omitting any factor), so that their sum is the middle term coefficient, giving the sign of the mid term to the larger group and the other sign  to the smaller, we have  -(3X3)and +(2X2X2)
That is -9 and 8 so that (-9+8) = -1. Hence the mid term -z = (-9z+8z)
The golden rule is: if the sign of the product is minus, and the  middle term (the term in z) if  minus,then give the sign of the mid term to the larger set and the other sign to the smaller 
Therefore (2) becomes
z^2-z-72 = 0 
z^2+(-9z+8z)-72 = 0
z^2-9z+8z-72 = 0      (by additive associativity)
(z^2-9z)+(8z-72) = 0   (by additive associativity)
z(z-9)+8(z-9) = 0
zp + 8p = 0 where p = (z-9)
p(z+8) = 0
(z-9)(z+8)=0
(z-9) = gives z= 9 and (z+8) = 0 gives z = -8
Answer: The two values for z are 9 and -8
Verification:
The given equation is (z-5)(z+4)=52
Putting z=9 in the above,
LHS= (9-5)X(9+4) = 4X13 = 52 =RHS
Therefore our value for z = 9 is right
Putting z=-8 in the above,
LHS= (-8-5)X(-8+4) = (-13)X(-4) = 52 =RHS
Therefore our value for z = -8 is right