Respuesta :

budwilkins

[tex] \frac{2 {y}^{2} - 7y - 15 }{y - 5} = 2y \\ remainder \: \frac{2 {y}^{2} - 7y - 15 }{ - (2{y }^{2} - 10y + 0)} \\ = 3y - 15[/tex]
[tex] \frac{3y - 15}{y - 5} = 3 \\ remainder \: \frac{3y - 15}{ - (3y - 15)} = 0[/tex]
So the quotient is : 2y + 3, with no remainder
JonHenderson55
The answer is:    " 2y + 3 " .
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The quotient is:  " 2y + 3 " .
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Explanation:
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(y − 5) * (    ?    )  = 2y² − 7y − 15 ; 

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Solve for " (   ?  ) " ;
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Let us examine:


" 2y²  / y = 2y " 

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So:  (y − 5) * (2y²  ... ? ...   )  = 2y² − 7y − 15 ; 

Examine the:  " -7y " ;  and the "-15" ; 
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Factors of:  " - 15" ;  that add up to "-7" ; 
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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:
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                                    + 3y − 10y  =  − 7y²  ; 
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→ And rewrite the expression:  

                                =  2y²  − 7y²  −  15 ; 
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So:  " (y − 5) * (2y + 3) = 2y² − 7y − 15 " . 
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The answer is:    " 2y + 3 " .
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The quotient is:  " 2y + 3 " .
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Now, suppose we had tried: "(2y − 3)" ; first:
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→  " (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 ;
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And we would rewrite the expression as:  

                                 =  " 2y²  − 13y + 15 " ; which means that "this answer" is INCORRECT ;  since:
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   " 2y²  − 13y + 15 "  [tex] \neq [/tex]    " 2y² − 7y − 15 " ; 
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  →  and as such:  " 2x − 3"; is NOT the "quotient" — and is NOT the correct answer.
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