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.
Im Stuc on this maybe vin can help you
Sorry i couldent help
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
it is wrong as it is (2x-5) not (2x+5)
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
i couldent help 
