You've uploaded an image with several algebraic expressions that you would like factored. I’ll factor each of the expressions one by one:

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8. $a^4 + a^2 + 1$

Factorization:
This is a quadratic in terms of $a^2$, but it doesn't factor nicely over the reals. So, it stays as:

$a^4 + a^2 + 1 = (a^2 + a + 1)(a^2 - a + 1)$

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9. $a^8 + a^4 + 1$

Factorization:
Again, this is a similar form to the previous expression but in powers of $a^4$:

$a^8 + a^4 + 1 = (a^4 + a^2 + 1)(a^4 - a^2 + 1)$

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10. $a^4 + 324$

Factorization:
This is a sum of squares:

$a^4 + 324 = (a^2 + 18)(a^2 - 18)$

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11. $a^4 + b^4$

Factorization:
This can be factored using the sum of squares formula:

$a^4 + b^4 = (a^2 + b^2)((a + b)^2 + (a - b)^2)$

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12. $a^6 - 1$

Factorization:
This is a difference of squares:

$a^6 - 1 = (a^3 - 1)(a^3 + 1)$ Both of these terms can be further factored:

$a^3 - 1 = (a - 1)(a^2 + a + 1)$ $a^3 + 1 = (a + 1)(a^2 - a + 1)$ Thus:

$a^6 - 1 = (a - 1)(a + 1)(a^2 + a + 1)(a^2 - a + 1)$

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13. $a^6 - 1$ (Same as above)
Factorization:
This is identical to #12, so the factorization remains:

$a^6 - 1 = (a - 1)(a + 1)(a^2 + a + 1)(a^2 - a + 1)$

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14. $a^{12} - 2a^6 + 1$

Factorization:
This is a perfect square trinomial:

$a^{12} - 2a^6 + 1 = (a^6 - 1)^2$

We already factored $a^6 - 1$, so:

$(a^6 - 1)^2 = ((a - 1)(a + 1)(a^2 + a + 1)(a^2 - a + 1))^2$

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15. $a^4 + 4a^2 - 5$

Factorization:
This is a quadratic in terms of $a^2$:

$a^4 + 4a^2 - 5 = (a^2 + 5)(a^2 - 1)$

Further factor $a^2 - 1$:

$a^2 - 1 = (a - 1)(a + 1)$ Thus:

$a^4 + 4a^2 - 5 = (a^2 + 5)(a - 1)(a + 1)$

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16. $a^4 + 5a^2 + 1$

Factorization:
This doesn't have a simple factorization over the reals, so it stays in its current form:

$a^4 + 5a^2 + 1 = (a^2 + \text{something complex})(a^2 + \text{something else complex})$

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17. $2a^4 + a^3 + 4a^2 + a + 2$

Factorization:
This can be tricky, but after factoring by grouping or trial, it factors as:

$2a^4 + a^3 + 4a^2 + a + 2 = (a + 1)(2a^3 - a^2 + 2)$

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18. $a^4 + 3a^3 + 4a^2 - 6a - 12$

Factorization:
This factors as:

$a^4 + 3a^3 + 4a^2 - 6a - 12 = (a^2 + 6)(a^2 - 1)$ Factor $a^2 - 1$:

$a^2 - 1 = (a - 1)(a + 1)$ Thus:

$a^4 + 3a^3 + 4a^2 - 6a - 12 = (a^2 + 6)(a - 1)(a + 1)$

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19. $a^3 + a^2 - a - 2$

Factorization:
Factoring by grouping:

$a^3 + a^2 - a - 2 = (a + 2)(a^2 - 1)$ Factor $a^2 - 1$:

$a^2 - 1 = (a - 1)(a + 1)$ Thus:

$a^3 + a^2 - a - 2 = (a + 2)(a - 1)(a + 1)$

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20. $2a^3 - a^2 + 3$

Factorization:
This is difficult to factor cleanly and does not have a simple factorization over the reals.

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That concludes the factorization of all the expressions. Would you like further details or explanations on any of the steps?

Here are some questions to expand your understanding:

1. What is the general formula for factoring differences of squares?
2. How do you recognize a perfect square trinomial for factoring?
3. Can you explain how to factor by grouping?
4. What is the method used to factor expressions that are quartic or higher in degree?
5. How does factoring help in solving polynomial equations?

Tip: Always check if the polynomial fits a known pattern (e.g., difference of squares, sum of cubes) before attempting other methods like trial and error.