The answer is: "
2y + 3 " .
____________________________________________________The quotient is: "
2y + 3 " .
____________________________________________________
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 "<span>±15" ;
</span> (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
<u>NOT</u> the "
quotient" — and is
<u>NOT</u><u> </u>the correct answer.
____________________________________________________