Factor the trinomial completely. $6 a ^ { 2 } - 6 a - 120$
A) $6 ( a + 5 ) ( a - 4 )$
B) $( 6 a + 5 ) ( a - 24 )$
C) $( 6 a - 30 ) ( a + 4 )$
D) $6 ( a - 5 ) ( a + 4 )$
E) not factorable

Question

Quizplus · Accepted Answer

First, factor out the greatest common factor, which is 6: $6(a^2 - a - 20)$. Then, factor the quadratic $a^2 - a - 20$ into $(a - 5)(a + 4)$. Thus, the complete factorization is $6(a - 5)(a + 4)$.

Factor the Trinomial Completely 6a2−6a−1206 a ^ { 2 } - 6 a - 1206a2−6a−120

Factor the Trinomial Completely $6 a ^ { 2 } - 6 a - 120$