The correct answer is:
Answer choice: [A]:__________________________________________________________→
"
" ; "
{ u 
± 3 } " ;
→ or, write as: "
u / (u − 3) " ;
{"
u ≠ 3 "
} AND:
{"
u ≠ -3 "
} ;
__________________________________________________________Explanation:__________________________________________________________ We are asked to simplify:
;
Note that the "numerator" —which is: "(u² + 3u)" — can be factored into:
→ " u(u + 3) " ;
And that the "denominator" —which is: "(u² − 9)" — can be factored into:
→ "(u − 3) (u + 3)" ;
___________________________________________________________Let us rewrite as:
___________________________________________________________→
;
___________________________________________________________→ We can simplify by "canceling out" BOTH the "
(u + 3)" values; in BOTH the "numerator" AND the "denominator" ; since:
"
" ;
→ And we have:
_________________________________________________________→ "
" ; that is: " u / (u − 3) " ; { u
} .
and: { u
} .
→ which is: "
Answer choice: [A] " .
_________________________________________________________NOTE: The "denominator" cannot equal "0" ; since one cannot "divide by "0" ;
and if the denominator is "(u − 3)" ; the denominator equals "0" when "u = -3" ; as such:
"u
3" ;
→ Note: To solve: "u + 3 = 0" ;
Subtract "3" from each side of the equation;
→ " u + 3 − 3 = 0 − 3 " ;
→ u = -3 (when the "denominator" equals "0") ;
→ As such: " u
-3 " ;
Furthermore, consider the initial (unsimplified) given expression:
→
;
Note: The denominator is: "
(u² − 9)" .
The "denominator" cannot be "0" ; because one cannot "divide" by "0" ;
As such, solve for values of "u" when the "denominator" equals "0" ; that is:
_______________________________________________________ → " u² − 9 = 0 " ;
→ Add "9" to each side of the equation ;
→ u² − 9 + 9 = 0 + 9 ;
→ u² = 9 ;
Take the square root of each side of the equation;
to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ;
→ √(u²) = √9 ;
→
| u
| = 3 ;
→ " u = 3" ; AND; "u = -3 " ;
We already have: "u = -3" (a value at which the "denominator equals "0") ;
We now have "u = 3" ; as a value at which the "denominator equals "0");
→ As such: "
u" ; "
u -3 " ;
or, write as: "
{ u ± 3 } " .
_________________________________________________________
