Factor completely. If the polynomial cannot be factored, write "prime."
-$x^{2}-14 x+48$
A) $(x+6)(x+8)$
B) $(x-6)(x-8)$
C) $(x-24)(x-2)$
D) Prime

Name: Factor completely. If the polynomial cannot be factored, write "prime." -$x^{2}-14 x+48$ A) $(x+6)(x+8)$ B) $(x-6)(x-8)$ C) $(x-24)(x-2)$ D) Prime
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Question

Quizplus · Accepted Answer

The correct factorization of $x^{2}-14x+48$ is found by looking for two numbers that multiply to 48 and add up to -14. Those numbers are -6 and -8, leading to the factorization $(x-6)(x-8)$.

Factor Completely x2−14x+48x^{2}-14 x+48x2−14x+48 A) (x+6)(x+8)(x+6)(x+8)(x+6)(x+8) B) (x−6)(x−8)(x-6)(x-8)(x−6)(x−8) C) (x−24)(x−2)(x-24)(x-2)(x−24)(x−2) D) Prime

Factor Completely $x^{2}-14 x+48$ A) $(x+6)(x+8)$ B) $(x-6)(x-8)$ C) $(x-24)(x-2)$ D) Prime