Factor completely.
-$5 x^{2}-45 x+100$
A) $(x-4)(5 x-25)$
B) $5(x-20)(x+1)$
C) $(5 x-20)(x-5)$
D) $5(x-4)(x-5)$

Name: Factor completely. -$5 x^{2}-45 x+100$ A) $(x-4)(5 x-25)$ B) $5(x-20)(x+1)$ C) $(5 x-20)(x-5)$ D) $5(x-4)(x-5)$
Brand: Quizplus
Rating: 4.7 (42 reviews)

Question

Quizplus · Accepted Answer

The expression can be factored by first taking out the greatest common factor, which is 5, resulting in $5(x^2 - 9x + 20)$. Then, the quadratic $x^2 - 9x + 20$ can be factored into $(x-4)(x-5)$. Thus, the complete factorization is $5(x-4)(x-5)$.

Factor Completely 5x2−45x+1005 x^{2}-45 x+1005x2−45x+100 A) (x−4)(5x−25)(x-4)(5 x-25)(x−4)(5x−25) B) 5(x−20)(x+1)5(x-20)(x+1)5(x−20)(x+1) C) (5x−20)(x−5)(5 x-20)(x-5)(5x−20)(x−5) D) 5(x−4)(x−5)5(x-4)(x-5)5(x−4)(x−5)

Factor Completely $5 x^{2}-45 x+100$ A) $(x-4)(5 x-25)$ B) $5(x-20)(x+1)$ C) $(5 x-20)(x-5)$ D) $5(x-4)(x-5)$