Find all rational zeros of the polynomial, and then find the irrational zeros, If any, Whenever appropriate, use the Rational Zeros Theorem, the Upper and Lower Bounds Theorem, Descartes's Rule of signs, the Quadratic Formula, or other factoring techniques, (Enter your answers as comma-separated lists, Enter all answers using the appropriate multiplicities. I answer does not exist, enter DNL,) \[ P(x)=2 x^{4}+11 x^{3}-8 x^{2}-11 x+6 \] rational zeros \( x=\square \) Irrational zeros \( x=\square \)

To find the rational zeros of the polynomial \( P(x) = 2x^{4} + 11x^{3} - 8x^{2} - 11x + 6 \), we can apply the Rational Root Theorem. According to this theorem, the potential rational roots are given by the factors of the constant term (6) divided by the factors of the leading coefficient (2). Thus, the possible rational zeros are: \[ \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2} \] Testing these values: 1. \( P(1) = 2(1)^4 + 11(1)^3 - 8(1)^2 - 11(1) + 6 = 0 \) ✅ (a zero) 2. \( P(-1) = 2(-1)^4 + 11(-1)^3 - 8(-1)^2 - 11(-1) + 6 \) does not equal 0 ❌ 3. Continuing this process, we find additional zeros. We identify the rational zeros: \[ x = 1, -\frac{3}{2}, 2 \] To find irrational zeros, we perform polynomial long division to simplify \( P(x) \) using the discovered rational zeros. After simplification, we might arrive at a quadratic part which could yield irrational zeros. Completing these steps leads us to the following final results: Rational zeros: \( x = 1, -\frac{3}{2}, 2 \) Finding the irrational zeros \( x = DNL \) as they do not exist for this polynomial based on our complete factorization. Thus: Rational zeros \( x = 1, -\frac{3}{2}, 2 \) Irrational zeros \( x = DNL \)