The answer is: "
2y + 3 " .
____________________________________________________The quotient is: "
2y + 3 " .
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Step-by-step explanation:_____________________________________________(y − 5) *
( ? ) = 2y² − 7y − 15 ;
_____________________________________________Solve for "
( ? ) " ;
_____________________________________________Let us examine:
" 2y² / y = 2y "
_______________________________________________So: (y − 5) *
(2y² ... ? ... ) = 2y² − 7y − 15 ;
Examine the: " -7y " ; and the "-15" ;
____________________________________________________Factors of: " - 15" ; that add up to "-7" ;
____________________________________________________Note: the "2" in the "2y² " ; Think of the factors of "
±15" ;
(15, 1, 5, 3)
We have "15 and 1" are out;
so "5 and 3" are the options;
we have a "5" (that is, a "-5") in "divisor).
Note the: "-15" ; and we have a "given divisor" containing "-5".
Note: " (-5)" * ( what?) = "(-15)" ? The answer is: " (-15) ÷ (-5) = (3).
The answer is: "3" ; {that is; "positive 3" . }.
Try: (y − 5) * (2y + 3) = ?
Note: (a + b) (c + d) = ac + ad + bc + bd ;
→ (y − 5) * (2y + 3) = (y * 2y) + (y * 3) + (-5 * 2y) + (-5 * 3) ;
= 2y² + 3y + (-10y) + (-15) ;
= 2y² + 3y − 10y − 15 ;
→ Combine the "like terms" ; as follows:
____________________________________________________
+ 3y − 10y = − 7y² ;
____________________________________________________
→ And rewrite the expression:
= 2y² − 7y² − 15 ;
____________________________________________________
So: " (y − 5) * (2y + 3) = 2y² − 7y − 15 " .
____________________________________________________
The answer is: " 2y + 3 " .
____________________________________________________
The quotient is: " 2y + 3 " .
____________________________________________________
Now, suppose we had tried: "(2y − 3)" ; first:
____________________________________________________
→ " (y − 5) (2y − 3) " = ? ;
Note: (a + b) (c + d) = ac + ad + bc + bd ;
→ " (y − 5) (2y − 3) = (y * 2y) + (y * -3) + (-5*2y) + (-5 * -3) ;
= ( 2y² ) + ( -3y ) + ( -10y ) + (15) ;
= 2y² − 3y − 10y + 15 ;
→ Combine the "like terms" ; as follows:
− 3y − 10y = − 13y ;
_____________________________________________________
And we would rewrite the expression as:
= " 2y² − 13y + 15 " ; which means that "this answer" is INCORRECT ; since:
_____________________________________________________
" 2y² − 13y + 15 "
" 2y² − 7y − 15 " ;
_____________________________________________________
→ and as such: " 2x − 3"; is NOT the "quotient" — and is NOT the correct answer.
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