Because all the attempts to factor this by breaking it up into the form #(Ax+B)(Cx+D)# fail, a good method to try is using the quadratic equation.

The given equation #y=4x^2+16x+9# is in standard form

#y=ax^2+bx+c#

with

#a=4#

#b=16#

#c=9#

The quadratic equation finds #x# for #4x^2+16x+9=0#

#x=(-b+-sqrt(b^2-4ac))/(2a)=(-16+-sqrt(16^2-4(4)(9)))/(2(4))#

#(-16+-sqrt(16^2-144))/(8)=(-16+-sqrt(112))/(8)=(-16+-4sqrt(7))/8#

The "#+-#" symbol means "both + and #-#", so this gives us that the two values of #x# that make #4x^2+16x+9=0# are

#x=(-16+4sqrt(7))/8# and #x=(-16-4sqrt(7))/8#

To create the final factorization of the original question, solve the two above equations for zero and multiply the two expressions:

#4x^2+16x+9=((8)/(4sqrt(7))x+16/(4sqrt(7)))(8/(-4sqrt(7))x+16/(-4sqrt(7)))#

The final step is technically optional, but teachers like it enough to grade you on it. Simplify:

#4x^2+16x+9=((2sqrt(7))/(7)x+(4sqrt(7))/(7))((-2sqrt(7))/(7)x-(4sqrt(7))/(7))#