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avatar+73 

7x^2  + nx - 11

 

Note  that  we have   

(7x  - 11)  (x + 1)      =    7x^2 - 11x + 7x  - 11   =   7x^2 - 4x - 11

(7x - 1)   (x + 11)     =     7x^2 -1x + 77x - 11    =    7x^2 + 76x - 11

(7x + 11) (x - 1)       =     7x^2  + 11x - 7x  - 11  =   7x^2 + 4x  -11

(7x + 1) (x - 11)   =         7x^2 - 77x  +1x - 11   =    7x^2  - 76x  - 11

 

So what are the actual possible values of $n$?

Also, another question that is apparently identicle to this you answered with :

7x^2  + nx  - 11

 

7 and 11 are prime....so the possible factorizations are

 

(7x + 11) ( x + 1)   ⇒   n =  18

 

(7x + 1) (x + 11)   ⇒  n = 78

 

(-7x + 1)(-x + 11)  ⇒ n = -78

 

(7x - 11)  ( x - 1) ⇒  n = -18

 

 

Big Confused,

MS

 Nov 19, 2020
 #1
avatar+116125 
+1

Note  that

 

I think that answer was for 7x^2  + nx  +11    which  isn't the same as   what you asked    =   7x^2 + nx  -11

 

In your question, n  has the values   -76, -4, 4 , 76  

 

cool cool cool  

 Nov 19, 2020
 #2
avatar+73 
0

So all the possible values of $n$ are $-76,-4,4,76?$

MathzSolver111  Nov 19, 2020
 #3
avatar+116125 
0

Yep......

 

cool cool cool

CPhill  Nov 19, 2020

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