Answer :
Let's factor the expression [tex]\(1024 a^2 - 400\)[/tex] step-by-step.
### Step 1: Identify a common structure
Observe that both terms, [tex]\(1024 a^2\)[/tex] and [tex]\(400\)[/tex], are perfect squares.
[tex]\[ 1024 a^2 = (32a)^2 \quad \text{and} \quad 400 = 20^2 \][/tex]
### Step 2: Apply the difference of squares formula
Recall the difference of squares formula:
[tex]\[ x^2 - y^2 = (x - y)(x + y) \][/tex]
In this case, we can set:
[tex]\[ x = 32a \quad \text{and} \quad y = 20 \][/tex]
### Step 3: Substitute and factor the expression
Using the difference of squares formula, we substitute [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ 1024 a^2 - 400 = (32a)^2 - 20^2 = (32a - 20)(32a + 20) \][/tex]
### Step 4: Verify the factors
To ensure our factors are correct, we can expand [tex]\( (32a - 20)(32a + 20) \)[/tex]:
[tex]\[ (32a - 20)(32a + 20) = 32a \cdot 32a + 32a \cdot 20 - 20 \cdot 32a - 20 \cdot 20 \][/tex]
[tex]\[ = 1024a^2 + 640a - 640a - 400 \][/tex]
[tex]\[ = 1024a^2 - 400 \][/tex]
This shows that our factorization is correct.
### Conclusion
[tex]\[ 1024 a^2 - 400 = (32a - 20)(32a + 20) \][/tex]
Hence, the correct factorization is:
[tex]\[ \boxed{(32a - 20)(32a + 20)} \][/tex]
### Step 1: Identify a common structure
Observe that both terms, [tex]\(1024 a^2\)[/tex] and [tex]\(400\)[/tex], are perfect squares.
[tex]\[ 1024 a^2 = (32a)^2 \quad \text{and} \quad 400 = 20^2 \][/tex]
### Step 2: Apply the difference of squares formula
Recall the difference of squares formula:
[tex]\[ x^2 - y^2 = (x - y)(x + y) \][/tex]
In this case, we can set:
[tex]\[ x = 32a \quad \text{and} \quad y = 20 \][/tex]
### Step 3: Substitute and factor the expression
Using the difference of squares formula, we substitute [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ 1024 a^2 - 400 = (32a)^2 - 20^2 = (32a - 20)(32a + 20) \][/tex]
### Step 4: Verify the factors
To ensure our factors are correct, we can expand [tex]\( (32a - 20)(32a + 20) \)[/tex]:
[tex]\[ (32a - 20)(32a + 20) = 32a \cdot 32a + 32a \cdot 20 - 20 \cdot 32a - 20 \cdot 20 \][/tex]
[tex]\[ = 1024a^2 + 640a - 640a - 400 \][/tex]
[tex]\[ = 1024a^2 - 400 \][/tex]
This shows that our factorization is correct.
### Conclusion
[tex]\[ 1024 a^2 - 400 = (32a - 20)(32a + 20) \][/tex]
Hence, the correct factorization is:
[tex]\[ \boxed{(32a - 20)(32a + 20)} \][/tex]