To factorize 8! — 5(7!), we can first write both terms with a common factor of 7!:
8! — 5(7!) = 7!(8/7) — 5(7!)
Simplifying further:
= 7! * (8/7 — 5)
= 7! * (8/7 – 5/1)
= 7! * (8/7 – 35/7)
= 7! * (8 – 35)/7
= 7! * (-27/7)
= 7! * (-3)
Thus, the factorization of 8! — 5(7!) is 7! * -3, which is equivalent to -3 * 7!.
The correct option is D. 3! x 7!.
Factorize 8! — 5(7!)
A. 4! 7!
B. 3 — 7!
C. 4 x 7!
D. 3! x 7!
3 answers
sorry - 8! = 8*7!, not 7! * 8/7, so you start with
8*7! - 5*7! = 3*7!
8*7! - 5*7! = 3*7!
Apologies for the mistake. You are correct.
To factorize 8! - 5(7!), we can first rewrite 8! as 8 * 7!:
8! - 5(7!) = 8(7!) - 5(7!)
Now we have a common factor of 7! in both terms. Factoring it out, we get:
= 7!(8 - 5)
= 7!(3)
Therefore, the factorization of 8! - 5(7!) is 3 * 7!.
The correct option is D. 3! x 7!.
To factorize 8! - 5(7!), we can first rewrite 8! as 8 * 7!:
8! - 5(7!) = 8(7!) - 5(7!)
Now we have a common factor of 7! in both terms. Factoring it out, we get:
= 7!(8 - 5)
= 7!(3)
Therefore, the factorization of 8! - 5(7!) is 3 * 7!.
The correct option is D. 3! x 7!.