Factor completely. Check by multiplying the factors. $16 x ^ { 2 } - 289$
A) $( 4 x + 17 ) ( 4 x + 17 )$
B) $( 4 x - 17 ) ( 4 x + 17 )$
C) $( x - 17 ) ( 4 x + 17 )$
D) $( 4 x - 17 ) ( 4 x + 4 )$
E) $( 17 x - 17 ) ( 4 x + 17 )$

Question

Quizplus · Accepted Answer

We can factor the expression using the difference of squares formula, which states that for any numbers a and b, $a^2-b^2 = (a+b)(a-b)$. In this case, we have:
$$16x^2 - 289 = (4x)^2 - 17^2 = (4x + 17)(4x - 17)$$
To check our answer, we can use FOIL (First, Outer, Inner, Last) method:
$$(4x + 17)(4x - 17) = 16x^2 - 68x + 68x - 289 = 16x^2 - 289$$
Therefore, our answer is correct and the best choice is (B).

Factor Completely 16x2−28916 x ^ { 2 } - 28916x2−289 A) (4x+17)(4x+17)( 4 x + 17 ) ( 4 x + 17 )(4x+17)(4x+17)

Factor Completely $16 x ^ { 2 } - 289$ A) $( 4 x + 17 ) ( 4 x + 17 )$