What are the factors of [tex]$x^2-64$[/tex]?
A. Prime
B. [tex]$(x-4)(x+16)$[/tex]
C. [tex]$(x-8)(x-8)$[/tex]
D. [tex]$(x+8)(x-8)$[/tex]
Answer (1)
Recognize the expression x 2 − 64 as a difference of squares.
Apply the difference of squares factorization: a 2 − b 2 = ( a − b ) ( a + b ) .
Identify a = x and b = 8 , so x 2 − 64 = ( x − 8 ) ( x + 8 ) .
The factors of x 2 − 64 are ( x − 8 ) ( x + 8 ) .
Explanation
Recognizing the Pattern We are asked to find the factors of the expression x 2 − 64 . This looks like a difference of squares, which has a special factoring pattern.
Applying the Difference of Squares Recall that the difference of squares can be factored as follows: a 2 − b 2 = ( a − b ) ( a + b ) . In our case, we have x 2 − 64 , which can be written as x 2 − 8 2 since 64 = 8 2 .
Finding the Factors Now we can apply the difference of squares factorization with a = x and b = 8 : x 2 − 64 = x 2 − 8 2 = ( x − 8 ) ( x + 8 ) . Therefore, the factors of x 2 − 64 are ( x − 8 ) and ( x + 8 ) .
Verifying the Answer Let's check the given options:
Prime: This is not a factorization. ( x − 4 ) ( x + 16 ) : Expanding this gives x 2 + 16 x − 4 x − 64 = x 2 + 12 x − 64 , which is not equal to x 2 − 64 .
( x − 8 ) ( x − 8 ) : Expanding this gives x 2 − 8 x − 8 x + 64 = x 2 − 16 x + 64 , which is not equal to x 2 − 64 .
( x + 8 ) ( x − 8 ) : Expanding this gives x 2 − 8 x + 8 x − 64 = x 2 − 64 , which is the correct factorization.
Examples
The difference of squares factorization is a useful tool in many areas of mathematics and physics. For example, consider a scenario where you need to calculate the area of a ring-shaped region. If the outer radius is x and the inner radius is 8, the area of the ring is given by π x 2 − π ( 8 2 ) = π ( x 2 − 64 ) . Using the factorization, we can rewrite this as π ( x − 8 ) ( x + 8 ) . This can simplify calculations or provide insights into the relationship between the radii and the area.
