$$ P(x)=x^{3}+6 x^{2}-32 $$ 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. $$ P(x)=\square $$

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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:
  $$x	imes x^{2}+x	imes 8x+x	imes 16-2x^{2}-2	imes 8x-2	imes 16$$
- step3: Factor the expression:
  $$x\left(x^{2}+8x+16ight)-2\left(x^{2}+8x+16ight)$$
- step4: Factor the expression:
  $$\left(x-2ight)\left(x^{2}+8x+16ight)$$
- step5: Factor the expression:
  $$\left(x-2ight)\left(x+4ight)^{2}$$
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 $$ P(x) = (x-2)(x+4)^{2} $$.
The zeros of the polynomial are $$ x = 2 $$ and $$ x = -4 $$. The factored form of the polynomial is $$ P(x) = (x-2)(x+4)^2 $$.

\( P(x)=x^{3}+6 x^{2}-32 \) 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. \( P(x)=\square \)

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