Let's analyze and solve each problem one by one.

Problem 14

The expression to factorize is: $a^5 + a^4 - 2a^3 - 2a^2 + a + 1$

To factorize, we can try grouping terms and look for common factors.

Grouping Terms

1. Group as $(a^5 + a^4) - (2a^3 + 2a^2) + (a + 1)$.
2. Factor out terms from each group:
   • From $(a^5 + a^4)$, factor out $a^4$ to get $a^4(a + 1)$.
   • From $(2a^3 + 2a^2)$, factor out $-2a^2$ to get $-2a^2(a + 1)$.
   • The last group $(a + 1)$ remains as is.

Thus, we get: $a^4(a + 1) - 2a^2(a + 1) + (a + 1)$

3. Now, factor out $(a + 1)$: $(a + 1)(a^4 - 2a^2 + 1)$

4. The expression inside the parentheses $(a^4 - 2a^2 + 1)$ can be written as a perfect square: $a^4 - 2a^2 + 1 = (a^2 - 1)^2$

5. Therefore, the factorization is: $(a + 1)(a^2 - 1)^2$

6. Notice that $a^2 - 1$ can be further factored as $(a + 1)(a - 1)$.

So the complete factorization is: $(a + 1)^2(a - 1)^2$

This matches option B: $(a + 1)^3 \cdot (a - 1)^2$.

Problem 15

The expression to factorize is: $(a^2 + 16)^2 - 64a^2$

This expression resembles the difference of squares, which is given by $A^2 - B^2 = (A + B)(A - B)$.

Applying Difference of Squares

1. Set $A = a^2 + 16$ and $B = 8a$.

2. Rewrite the expression as: $(a^2 + 16)^2 - (8a)^2$

3. Now apply the difference of squares formula: $((a^2 + 16) + 8a)((a^2 + 16) - 8a)$

4. Simplify each factor:

   • $(a^2 + 16 + 8a) = (a^2 + 8a + 16) = (a + 4)^2$
   • $(a^2 + 16 - 8a) = (a^2 - 8a + 16) = (a - 4)^2$

So the factorization is: $(a - 4)^2(a + 4)^2$

This matches option C: $(a - 4)^2 \cdot (a + 4)^2$.

Summary of Answers

• Problem 14 Answer: B) $(a + 1)^3 \cdot (a - 1)^2$
• Problem 15 Answer: C) $(a - 4)^2 \cdot (a + 4)^2$

Would you like further details on the steps?

Here are some related questions you might consider:

1. How does the difference of squares formula work?
2. What are alternative methods for factoring expressions?
3. How do you recognize when to use grouping in factorization?
4. What are some special products that frequently appear in algebra?
5. How can complex polynomials be factored using synthetic division?

Tip: Always check if an expression can be simplified with identities like difference of squares or perfect square trinomials for faster factoring.