Find all the real zeros of the polynomial. Use the quadratic formula if necessary. (Enter your answers as a comma-separated list. Enter all answers including repetitions. \[ P(x)=3 x^{3}+18 x^{2}+21 x-6 \] \( x=-2,-1 \sqrt{5} \) You do not have the correct number of terms in your comma-separated list.

To find the real zeros of the polynomial \( P(x) = 3x^3 + 18x^2 + 21x - 6 \), we can begin by factoring or using the Rational Root Theorem. However, we can also check for potential rational roots like \( x = -2 \) and \( x = -1 \). Let's synthetically divide \( P(x) \) by \( x + 2 \) and \( x + 1 \), which you suspect to be roots: 1. **For \( x = -2 \)**: - Synthetic division gives us \( P(-2) = 0 \), so \( x + 2 \) is indeed a factor. The quotient polynomial will be \( 3x^2 + 12x + 3 \). 2. **Now we factor \( 3x^2 + 12x + 3 \)**: - This simplifies to \( x^2 + 4x + 1 \) after factoring out 3. We can apply the quadratic formula here: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \Rightarrow x = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}. \] Combining all the roots, we have \( x = -2, -1, -2 + \sqrt{3}, -2 - \sqrt{3} \). Make sure to express your answer with all roots shown: \[ x = -2, -1, -2 + \sqrt{3}, -2 - \sqrt{3}. \] Pulling from the world of mathematics, did you know that polynomials can have as many roots as their highest degree? For a cubic polynomial like this, there can be up to three real roots—hence, the various values we've found! For a more advanced dive, you might want to explore how polynomial division works or the Fundamental Theorem of Algebra. It’s fascinating how these concepts create connections between simple roots and complex polynomials!