Answer:
[tex]\large\boxed{x=-\dfrac{1}{3}}[/tex]
Step-by-step explanation:
[tex]f(x)=9x^2+6x+1\\\\\text{The zeros are for}\ f(x)=0\\\\9x^2+6x+1=0\\\\9x^2+3x+3x+1=0\\\\3x(3x+1)+1(3x+1)=0\\\\(3x+1)(3x+1)=0\\\\(3x+1)^2=0\iff3x+1=0\qquad\text{subtract 1 from both sides}\\\\3x=-1\qquad\text{divide both sides by 3}\\\\x=-\dfrac{1}{3}[/tex]
