All the real zeros of the given polynomial are integers. Find the zeros. (Enter your answers as a comma-separated list. Enter all answers using the appropriate multiplicities.) \[ P(x)=x^{3}-2 x^{2}-13 x-10 \] Write the polynomial in factored form. \( P(x)=\square \)

To solve for the real zeros of   P(x) = x³ - 2x² - 13x - 10, we first look for integer zeros among the divisors of -10, which are ±1, ±2, ±5, ±10. Step 1. Test possible zeros:   P(-1) = (-1)³ - 2(-1)² - 13(-1) - 10       = -1 - 2 + 13 - 10       = 0. So, x = -1 is a zero, meaning (x + 1) is a factor. Step 2. Divide P(x) by (x + 1) using synthetic division:   Coefficients: 1, -2, -13, -10.     Carry down the 1.     Multiply: (-1) × 1 = -1; then add to -2 to get -3.     Multiply: (-1) × -3 = 3; then add to -13 to get -10.     Multiply: (-1) × -10 = 10; then add to -10 to get 0. The quotient is x² - 3x - 10. Step 3. Factor the quadratic:   x² - 3x - 10 factors as (x - 5)(x + 2) because (-5) + 2 = -3 and (-5) × 2 = -10. Thus, the complete factorization of P(x) is:   P(x) = (x + 1)(x - 5)(x + 2). The zeros of P(x) are x = -1, x = 5, and x = -2 (each with multiplicity 1). Answer: Zeros: -2, -1, 5 P(x) = (x + 1)(x - 5)(x + 2)