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`14,`

`a,` Ta có :

$\left.\begin{array}{l}5x^3+14x^2+12x-5\\  \quad\quad\quad\:4x^2+12x\\\:\:\quad\quad\quad\quad\quad\quad4x-5\\\:\:\:\:\:\: \quad\quad\quad\quad\quad\quad\:-13\\\:\:\:\:\:\:\end{array}\right|\!\!\!\;\!\!\begin{array}{l}\dfrac{\ \ x+2\ \ }{\ \ 5x^2+4x+4\ \ }\\\\\\\end{array}$

`=> Q = 5x^2 + 4x + 4 ; R = -13 ; B = x+2`

Vậy `A = (x+2).(5x^2+4x+4) - 13`

`b,` Ta có :

$\left.\begin{array}{l}-\dfrac{3}{2}x^3+5x^2-7x+11\\  \quad\quad\quad\:\:\:\:3x^2-7x\\\:\:\quad\quad\quad\quad\quad\:-3x+11\\\:\:\:\:\:\: \quad\quad\quad\quad\quad\quad\quad\:\:\:7\\\:\:\:\:\:\:\end{array}\right|\!\!\!\;\!\!\begin{array}{l}\dfrac{\\ -3x+4\ \ }{\ \ \dfrac{1}{2}x^2-x+1\ \ }\\\\\\\end{array}$

`=> Q =1/2 x^2 - x + 1 ; R = 7 ; B = -3x+4`

Vậy `A = (-3x+4).(1/2 x^2-  x+ 1) + 7`

`15,`

`a,` Để đa thức `A` chia hết cho đa thức `B` thì : 

`A(3) = 0`

`=> 3^3 - 2.3^2 + (a-3).3 - a - 1 = 0`

`=> 27 - 18 + 3a - 9 - a - 1 = 0`

`=> 2a = 1`

`=> a = 1/2`

Vậy `a = 1/2` để `A vdots B`

`b,` Ta có :

$\left.\begin{array}{l}2x^3+10x^2+(a-1)x+b-a\\  2x^3\quad\quad\quad+\quad\quad\:\;\:5x\\\:\:\quad\quad10x^2+(a-6)x+(b-a)\\\:\:\:\:\:\: \quad10x^2\quad\quad\quad\quad\quad+\quad25\\\quad\quad\quad\:(a-6)x+b-a-25\end{array}\right|\!\!\!\;\!\!\begin{array}{l}\dfrac{\ \ 2x^2+5\ \ }{\ \ x+5\ \ }\\\\\\\end{array}$

Để `A vdots B` thì số dư phải bằng `0`

`=> (a-6)x + b - a - 25 = 0`

`=> {(a-6=0),(b-a-25=0):}`

`=> {(a=6),(b=31):}`

Vậy `a=6 ; b= 31` để `A vdots B`