What is the solution set to the equation 0=(3a+1)^2(a-4) A) (1)/(3),-4 B) -(1)/(3),4 C) -(1)/(3),(1)/(3),-4 D) -(1)/(3),(1)/(3),4

To find the solution set to the equation $0=(3a+1)^{2}(a-4)$, we need to find the values of $a$ that make the equation equal to zero.<br /><br />The equation is a product of two factors: $(3a+1)^{2}$ and $(a-4)$. For the equation to be equal to zero, at least one of the factors must be equal to zero.<br /><br />Let's consider the first factor, $(3a+1)^{2}$. This factor is a perfect square, so it is always non-negative. The only way for it to be equal to zero is if $3a+1=0$. Solving this equation, we get $a=-\frac{1}{3}$.<br /><br />Now let's consider the second factor, $(a-4)$. This factor is a linear term, so it can be equal to zero if $a=4$.<br /><br />Therefore, the solution set to the equation is $\{ -\frac {1}{3},4\} $.<br /><br />So, the correct answer is B) $\{ -\frac {1}{3},4\} $.