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))
