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in the expansion of (2x-5)(3x^2+px+6)

the coefficient of the x term is half the coefficient of the x^2 term.

calculate the value of the constant p.

 Apr 14, 2022
 #1
avatar+122 
-2

Im Stuc on this maybe vin can help you

 Apr 14, 2022
 #2
avatar+122 
-2

Sorrysad i couldent help 

 Apr 14, 2022
 #3
avatar+23183 
+1

Expanding  (2x + 5)(3x2 + px + 6)  

                     =  6x3 + 2px2 + 12x

                                +15x2 + 5px + 30

                     =  6x3 + (2p + 15)2 + (5p + 12)x + 30

 

Since the coefficient of the x term is half the coefficient of the x2 term:

                      5p + 12  =  ½(2p + 15)

                  2(5p + 12)  =  2p + 15

                    10p + 24  =  2p + 15

                              8p  =  -9

                                p  =  -9/8

 Apr 14, 2022
 #4
avatar
0

it is wrong as it is (2x-5) not (2x+5)

Guest Apr 14, 2022
 #5
avatar+122390 
+1

Just a slight correction to  geno's answer

 

(2x  -5) (3x^2 + px + 6)  =

 

6x^3 - 15x^2 + 2px^2 - 5px + 12x -30  =

 

6x^3 + ( 2p - 15)x^2 + ( 12 - 5p)x  - 30

 

(12 - 5p)  = (1/2)(2p  -15)

 

2 ( 12 - 5p) = 2p -15

 

24 - 10p  = 2p  -15

 

24 + 15  =  12p

 

39 = 12p

 

p = 39/12  =  13/4

 

cool cool cool

 Apr 14, 2022

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