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Let $K$ be the splitting field of the polynomial $(x^3 + x - 1)(x^4 + x - 1)$ over $\mathbb{F}_3$. How many elements does $K$ contain? What I've already done is factoring $(x^3 + x - 1)(x^4 + x - 1)$ as $(1 + x)(x^2 - x - 1)(x^4 + x - 1)$, which allows me to conclude that the splitting field of $(x^3 + x - 1)(x^4 + x - 1)$ over $\mathbb{F}_3$ is the same as the splitting field of $(1 + x)(x^2 - x - 1)(x^4 + x - 1)$ over $\mathbb{F}_3$ which is the splitting field of $(x^2 - x - 1)(x^4 + x - 1)$ over $\mathbb{F}_3$. If we look at the splitting field of $(x^2 - x - 1)$ over $\mathbb{F}_3$ then we see that its simply $\mathbb{F}_9$. How do I continue with this?

EDIT; I think I've found the answer. Consider the splitting field of $(x^4 + x - 1)$ over $\mathbb{F}_3$. Since $(x^4 + x - 1)$ is irreducible over $\mathbb{F}_3$ it follows that this splitting field is just $\mathbb{F}_{3^4}$. Since $\mathbb{F}_{3^2}$ is a subfield of $\mathbb{F}_{3^4}$, we see that the splitting field of $(x^2 - x - 1)$ over $\mathbb{F}_3$ is already contained in the splitting field of $(x^4 + x - 1)$ over $\mathbb{F}_3$. Can anyone confirm if this is right or not?

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The OP has added an answer as an edit to the question, so I am copying it here to mark this question as answered.

Consider the splitting field of $(x^4 + x - 1)$ over $\mathbb{F}_3$. Since $(x^4 + x - 1)$ is irreducible over $\mathbb{F}_3$ it follows that this splitting field is just $\mathbb{F}_{3^4}$. Since $\mathbb{F}_{3^2}$ is a subfield of $\mathbb{F}_{3^4}$, we see that the splitting field of $(x^2 - x - 1)$ over $\mathbb{F}_3$ is already contained in the splitting field of $(x^4 + x - 1)$ over $\mathbb{F}_3$. Hence, the splitting field of $(x^3 + x - 1)(x^4 + x - 1)$ over $\mathbb{F}_3$ is $K = \mathbb{F}_{3^4}$.