Factor completely.
-$16 x ^ { 2 } + 49$
A) $( 4 x + 7 ) ^ { 2 }$
B) $( 4 x - 7 ) ^ { 2 }$
C) $( 4 x + 7 ) ( 4 x - 7 )$
D) prime

Name: Factor completely. -$16 x ^ { 2 } + 49$ A) $( 4 x + 7 ) ^ { 2 }$ B) $( 4 x - 7 ) ^ { 2 }$ C) $( 4 x + 7 ) ( 4 x - 7 )$ D) prime
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Question

Quizplus · Accepted Answer

The expression $16x^2 + 49$ is a sum of squares, which cannot be factored over the real numbers, making it prime.

Factor Completely 16x2+4916 x ^ { 2 } + 4916x2+49 A) (4x+7)2( 4 x + 7 ) ^ { 2 }(4x+7)2

Factor Completely $16 x ^ { 2 } + 49$ A) $( 4 x + 7 ) ^ { 2 }$