answer: 6 √35 ; 6* sqrt 35
step-by-step explanation:
√(7! ) =
√(7 * 6 * 5 *4 * 3 * 2 * 1 ) ;
→ {since: " 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 " .}.
= √7 *√6 * √5*√4*√3*√2*√1 = ?
note: "√1 = 1 " ; and any value, multiplied by "1" , results in that same value;
so we can eliminate the "√1" ;
the "√4" can be simplied to "2" ; since it is a "perfect square:
note of the other radicands are "perfect squares" .
so, we have:
√(7! ) = √(7 * 6 * 5 *4 * 3 * 2 * 1 ) ;
= 2 * √(7 * 6 * 5 * 3 * 2 ) ;
now, let us multiply: "(7 * 6 * 5 * 3 * 2) " ;
→ 7 * 6 * 5 * 3 * 2 = 1260 ;
as such: √(7 * 6 * 5 * 3 * 2 ) = √(1260) ;
now, can the radicand:
"1260" ; be factored —
[note: hereinafter, by "factor(s) //factor/factoring" , we will assume "factorable/factor(s)/factoring" as "regards to "non-zero, positive integers"] —into any perfect squares? ;
yes;
→ "100" is a "perfect square" since: " √100 = 10 " ; and: ↔ " 10² = 100 " .
we now that "100" is a factor of 1260; since "1260" is a non-zero positive integer of three (3) or more digits that ends in zero.
since: " 1260 = 126 * 100 " ; →