If we find a value #r# such that #2r^2+11r+6=0#, then we can say that the binomial #z-r# is a factor of #2z^2+11z+6#. In this case it turns out that there are two values of #r# that meet this criteria, however neither of them are rational. I am going to assume that you do not know the quadratic formula and try to explain how to arrive at the two values of #r# by completing the square.
We want to find out when
#2z^2+11z+6=0#.
Divide both sides of this equation by 2.
#z^2+11/2z+3=0#
Add 121/16 to both sides of this equation.
#z^2+11/2z+121/16+3=121/16#
Factor the first three terms of the left-hand side of this equation.
#(z+11/4)^2+3=121/16#
Subtract three from both sides of this equation.
#(z+11/4)^2=121/16-3#
Simplify.
#(z+11/4)^2=73/16#
Take the square root of both sides of this equation.
#z+11/4=pmsqrt(73/16)#
Subtract #11/4# from both sides of this equation.
#z=-11/4pmsqrt(73/16)#
Simplify
#z=(-11pmsqrt(73))/4#
This means that #(z-(-11+sqrt(73))/4)# and #(z-(-11-sqrt(73))/4)# are both factors of #2z^2+11z+6#. Let's evaluate
#(z-(-11+sqrt(73))/4)(z-(-11-sqrt(73))/4)#
#z^2-(-11+sqrt(73))/4z-(-11-sqrt(73))/4z+((-11+sqrt(73))(-11-sqrt(73)))/16#
#z^2+11/2z+(121-73)/16#
#z^2+11/2z+3#
Note that this is exactly 1/2 of our original expression. This means that the remaining factor is 2, and the fully-factored form is
#2(z-(-11+sqrt(73))/4)(z-(-11-sqrt(73))/4)#.
