The problem involves factoring the polynomial $64 - a^2 - 10ab - 25b^2$. Let's solve it step-by-step.

Step 1: Arrange the terms in standard form

The given polynomial is: $64 - a^2 - 10ab - 25b^2$ Rewriting it in standard order: $-a^2 - 10ab - 25b^2 + 64$

Step 2: Factor out a negative sign for clarity

Factoring out $-1$: $-(a^2 + 10ab + 25b^2 - 64)$

Step 3: Group terms for factoring

The expression inside the parentheses is: $a^2 + 10ab + 25b^2 - 64$ Notice the quadratic trinomial $a^2 + 10ab + 25b^2$ can be factored: $a^2 + 10ab + 25b^2 = (a + 5b)^2$ Substituting this back gives: $-( (a + 5b)^2 - 64 )$

Step 4: Apply the difference of squares

The expression $(a + 5b)^2 - 64$ is a difference of squares: $(A^2 - B^2) = (A - B)(A + B)$ Here, $A = (a + 5b)$ and $B = 8$, so: $(a + 5b)^2 - 64 = \big((a + 5b) - 8\big)\big((a + 5b) + 8\big)$

Step 5: Simplify the factors

Simplify the two terms: $(a + 5b) - 8 = a + 5b - 8$ $(a + 5b) + 8 = a + 5b + 8$

Final Answer

The fully factored form of the polynomial is: $-(a + 5b - 8)(a + 5b + 8)$

Would you like a detailed explanation of any part of the process?

Related Questions

1. What is the difference of squares, and how is it applied in polynomial factoring?
2. How can you recognize and factor perfect square trinomials like $a^2 + 10ab + 25b^2$?
3. Why is factoring out a negative sign useful in polynomial manipulations?
4. How can you generalize the process of factoring $A^2 - B^2$?
5. What are other techniques for factoring complex polynomials?

Tip

Always rearrange polynomials into standard form ($ax^2 + bx + c$) before attempting to factor or apply special formulas.