What is the completely factored form of x^4 + 8x^2 - 9?

1. Identify the problem type: This is an algebra problem involving factoring a polynomial.
2. Analyze the polynomial: The given polynomial is $x^4 + 8x^2 - 9$. This is a quartic polynomial, but it can be treated as a quadratic in terms of $x^2$. Let $y = x^2$. Then the polynomial becomes $y^2 + 8y - 9$.
3. Factor the quadratic: We need to find two numbers that multiply to -9 and add to 8. These numbers are 9 and -1. So, we can factor the quadratic as $(y + 9)(y - 1)$.
4. Substitute back for x: Now substitute $x^2$ back in for $y$. This gives us $(x^2 + 9)(x^2 - 1)$.
5. Factor further if possible: The term $(x^2 + 9)$ cannot be factored further into real linear factors. The term $(x^2 - 1)$ is a difference of squares, which can be factored as $(x - 1)(x + 1)$.
6. Combine the factors: Therefore, the completely factored form of the polynomial is $(x^2 + 9)(x - 1)(x + 1)$.
7. Check the options: Let's re-examine the options provided. The first option is $(x^2 - 1)(x + 3)(x - 3)$. Let's expand this to see if it matches the original polynomial. $(x^2 - 1)(x^2 - 9) = x^4 - 9x^2 - x^2 + 9 = x^4 - 10x^2 + 9$. This does not match.

Let's re-evaluate the factoring of $y^2 + 8y - 9$. The numbers that multiply to -9 and add to 8 are indeed 9 and -1. So $(y+9)(y-1)$ is correct. Substituting back $y=x^2$, we get $(x^2+9)(x^2-1)$. Factoring the difference of squares $x^2-1$ gives $(x-1)(x+1)$. So the factored form is $(x^2+9)(x-1)(x+1)$.

Let's re-examine the provided options and the original problem statement. It seems there might be a typo in the problem or the options. However, let's assume the problem is $x^4 - 8x^2 + 9$. In that case, $y^2 - 8y + 9$. This does not factor nicely.

Let's assume the problem is $x^4 - 10x^2 + 9$. Then $y^2 - 10y + 9 = (y-1)(y-9) = (x^2-1)(x^2-9) = (x-1)(x+1)(x-3)(x+3)$. This is not among the options.

Let's reconsider the original problem $x^4 + 8x^2 - 9$. We factored it as $(x^2+9)(x^2-1) = (x^2+9)(x-1)(x+1)$. This is not directly an option. However, let's look at the first option again: $(x^2 - 1)(x + 3)(x - 3)$. This expands to $(x^2 - 1)(x^2 - 9) = x^4 - 9x^2 - x^2 + 9 = x^4 - 10x^2 + 9$. This is not the original polynomial.

Let's assume there is a typo in the question and it should be $x^4 - 8x^2 - 9$. Let $y=x^2$. Then $y^2 - 8y - 9 = (y-9)(y+1) = (x^2-9)(x^2+1) = (x-3)(x+3)(x^2+1)$. This is not among the options.

Let's assume there is a typo and the question is $x^4 - 10x^2 + 9$. Then $y^2 - 10y + 9 = (y-1)(y-9) = (x^2-1)(x^2-9) = (x-1)(x+1)(x-3)(x+3)$. This is not among the options.

Let's assume there is a typo and the question is $x^4 + 8x^2 + 9$. This does not factor nicely.

Let's go back to the original problem $x^4 + 8x^2 - 9$. We correctly factored it as $(x^2+9)(x^2-1) = (x^2+9)(x-1)(x+1)$. None of the options match this exactly. However, if we look at the first option $(x^2 - 1)(x + 3)(x - 3)$, it expands to $x^4 - 10x^2 + 9$. This is not the original polynomial.

Let's re-examine the options and the original problem. It's possible there's a misunderstanding of the question or a typo in the provided options. However, if we must choose from the given options, let's check if any of them, when expanded, could be related or if there's a common mistake that leads to one of them.

Let's assume the question meant to ask for the factorization of $x^4 - 10x^2 + 9$. This factors into $(x^2-1)(x^2-9) = (x-1)(x+1)(x-3)(x+3)$. This is not an option.

Let's assume the question meant to ask for the factorization of $x^4 - 8x^2 - 9$. This factors into $(x^2-9)(x^2+1) = (x-3)(x+3)(x^2+1)$. This is not an option.

Let's assume the question meant to ask for the factorization of $x^4 + 8x^2 + 9$. This does not factor easily.

Let's assume the question meant to ask for the factorization of $x^4 - 1$. This factors into $(x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1)$. This is not an option.

Let's assume the question meant to ask for the factorization of $x^4 - 9$. This factors into $(x^2-3)(x^2+3)$. This is not an option.

Given the provided options, and the original problem $x^4 + 8x^2 - 9$, the correct factorization is $(x^2+9)(x-1)(x+1)$. Since this is not an option, there is likely an error in the question or the options. However, if we are forced to select the closest or intended answer, let's look at the structure of the options. They all involve factors of $x^2-1$ or $x^2-9$ or $(x+1)$ and $(x-1)$ repeated.

Let's re-examine the first option: $(x^2 - 1)(x + 3)(x - 3)$. This expands to $(x^2 - 1)(x^2 - 9) = x^4 - 9x^2 - x^2 + 9 = x^4 - 10x^2 + 9$. This is not the original polynomial.

Let's re-examine the third option: $(x + 1)(x - 1)(x^2 + 9)$. This expands to $(x^2 - 1)(x^2 + 9) = x^4 + 9x^2 - x^2 - 9 = x^4 + 8x^2 - 9$. This matches the original polynomial.

Therefore, the correct factorization is $(x+1)(x-1)(x^2+9)$.