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.) \[ P(x)=x^{3}-13 x^{2}+35 x+49 \] You do not have the correct number of terms in your comma-separated list. Write the polynomial in factored form. \( P(x)=(x-1)(x-6)(x-11) \) Need Help? Read li

We are given the polynomial   P(x) = x³ – 13x² + 35x + 49. Step 1. Find a zero by testing small integer values. Trying x = –1:   P(–1) = (–1)³ – 13(–1)² + 35(–1) + 49        = –1 – 13 – 35 + 49 = 0. Since P(–1) = 0, x = –1 is a root. Step 2. Factor P(x) by dividing out the factor (x + 1). Divide the polynomial by (x + 1) using synthetic division:   Coefficients: 1  –13  35  49   Using –1:   • Bring down 1.   • Multiply 1 by –1 to get –1; add to –13 to get –14.   • Multiply –14 by –1 to get 14; add to 35 to get 49.   • Multiply 49 by –1 to get –49; add to 49 to get 0. The quotient is x² – 14x + 49. Step 3. Factor the quadratic:   x² – 14x + 49 factors as (x – 7)²   (since 7 · 7 = 49 and 7 + 7 = 14). Thus, the fully factored form is   P(x) = (x + 1)(x – 7)². Step 4. Identify the zeros (including their multiplicities):   The zeros are x = –1 and x = 7 (with multiplicity 2). Final Answer:  Zeros: –1, 7, 7  Factored Form: (x + 1)(x – 7)².