Find all the zeros, real and nonreal, of the polynomial and use that information to express $p(x)$ as a product of linear factors. \[ p(x)=3 x^{2}-8 x+4 \] \[ p(x)= \]

Step 1 ：First, we set the polynomial equal to zero: \(3x^2 - 8x + 4 = 0\).

Step 2 ：Next, we can use the quadratic formula to find the roots of the polynomial. The quadratic formula is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation.

Step 3 ：Substituting the coefficients of our polynomial into the quadratic formula, we get \(x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4*3*4}}{2*3}\).

Step 4 ：Simplifying the expression under the square root, we get \(x = \frac{8 \pm \sqrt{64 - 48}}{6}\).

Step 5 ：Further simplifying, we get \(x = \frac{8 \pm \sqrt{16}}{6}\).

Step 6 ：Taking the square root of 16, we get \(x = \frac{8 \pm 4}{6}\).

Step 7 ：This gives us two solutions: \(x = \frac{8 + 4}{6} = 2\) and \(x = \frac{8 - 4}{6} = \frac{2}{3}\).

Step 8 ：So, the zeros of the polynomial are \(x = 2\) and \(x = \frac{2}{3}\).

Step 9 ：Finally, we can express the polynomial as a product of linear factors by factoring out the leading coefficient and using the zeros we found: \(p(x) = 3(x - 2)(x - \frac{2}{3})\).

Step 10 ：Checking our work, we can expand this product to verify that it equals the original polynomial: \(3(x - 2)(x - \frac{2}{3}) = 3x^2 - 8x + 4\).

Step 11 ：So, the polynomial \(p(x) = 3x^2 - 8x + 4\) can be expressed as a product of linear factors as \(\boxed{p(x) = 3(x - 2)(x - \frac{2}{3})}\).