text: # What are all the possible rational zeros for y=4x^3+12x^2+x+3 and how do you find all zeros?
  
  Nov 8, 2016
  
  Use the rational root theorem to find "possible" rational zeros:
  
  $\pm \frac{1}{4} , \pm \frac{1}{2} , \pm \frac{3}{4} , \pm 1 , \pm \frac{3}{2} , \pm 3$
  
  Factor by grouping to find the actual zeros: $- 3$, $\pm \frac{1}{2} i$
  
  #### Explanation:
  
  $f \left(x\right) = 4 {x}^{3} + 12 {x}^{2} + x + 3$
  
  By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $3$ and $q$ a divisor of the coefficient $4$ of the leading term.
  
  That means that the only possible rational zeros are:
  
  $\pm \frac{1}{4} , \pm \frac{1}{2} , \pm \frac{3}{4} , \pm 1 , \pm \frac{3}{2} , \pm 3$
  
  In fact, since all of the coefficients of $f \left(x\right)$ are positive, any Real zeros are negative, leaving us with possible rational zeros:
  
  $- \frac{1}{4} , - \frac{1}{2} , - \frac{3}{4} , - 1 , - \frac{3}{2} , - 3$
  
  We could try each of these in turn, but there's something else we can notice about this particular cubic: The ratio between the first and second terms is the same as that between the third and fourth terms. As a result, this cubic will factor by grouping:
  
  $4 {x}^{3} + 12 {x}^{2} + x + 3 = \left(4 {x}^{3} + 12 {x}^{2}\right) + \left(x + 3\right)$
  
  $\textcolor{w h i t e}{4 {x}^{3} + 12 {x}^{2} + x + 3} = 4 {x}^{2} \left(x + 3\right) + 1 \left(x + 3\right)$
  
  $\textcolor{w h i t e}{4 {x}^{3} + 12 {x}^{2} + x + 3} = \left(4 {x}^{2} + 1\right) \left(x + 3\right)$
  
  Hence $f \left(x\right)$ has one Real zero $x = - 3$
  
  The other two zeros are non-Real Complex since $4 {x}^{2} + 1 > 0$ for all Real values of $x$. We can factor this expression using the difference of squares identity:
  
  ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
  
  with $a = 2 x$ and $b = i$ as follows:
  
  $4 {x}^{2} + 1 = {\left(2 x\right)}^{2} - {i}^{2} = \left(2 x - i\right) \left(2 x + i\right)$
  
  Hence the two remaining zeros are $x = \pm \frac{1}{2} i$