\( P(x)=x^{3}+6 x^{2}-32 \) All the real zeros of the given polynomlal are integers. Find the zeros. (Enter your answers as a comma-separated list. Enter all answers including repetitions.) Write the polynomial in factored form. \( P(x)=\square \)

To find the zeros of the polynomial \( P(x) = x^{3} + 6x^{2} - 32 \), we need to solve the equation \( P(x) = 0 \). Given that all the real zeros of the polynomial are integers, we can start by factoring the polynomial to find its zeros. Let's factor the polynomial \( P(x) = x^{3} + 6x^{2} - 32 \) to find its zeros. Factor the expression by following steps: - step0: Factor: \(x^{3}+6x^{2}-32\) - step1: Calculate: \(x^{3}+8x^{2}+16x-2x^{2}-16x-32\) - step2: Rewrite the expression: \(x\times x^{2}+x\times 8x+x\times 16-2x^{2}-2\times 8x-2\times 16\) - step3: Factor the expression: \(x\left(x^{2}+8x+16\right)-2\left(x^{2}+8x+16\right)\) - step4: Factor the expression: \(\left(x-2\right)\left(x^{2}+8x+16\right)\) - step5: Factor the expression: \(\left(x-2\right)\left(x+4\right)^{2}\) The factored form of the polynomial \( P(x) = x^{3} + 6x^{2} - 32 \) is \( (x-2)(x+4)^{2} \). Now, we can find the zeros of the polynomial by setting each factor equal to zero and solving for \( x \). Setting \( x - 2 = 0 \), we find \( x = 2 \). Setting \( x + 4 = 0 \), we find \( x = -4 \). Therefore, the zeros of the polynomial are \( x = 2 \) and \( x = -4 \). The factored form of the polynomial is \( P(x) = (x-2)(x+4)^{2} \).