(x+m)(x-3)+7=(x+a)(x+b)

=>\(x^2-3x+mx-3m+7=x^2+\left(a+b\right)x+ab\)

=>\(x^2+x\left(m-3\right)-3m+7=x^2+\left(a+b\right)x+ab\)

=>m-3=a+b và -3m+7=ab

=>3m-9=3a+3b và -3m+7=ab

=>3m-9-3m+7=3a+3b+ab

=>ab+3a+3b=-2

=>a(b+3)+3b+9=-2+9

=>(a+3)(b+3)=7

=>(a+3;b+3)∈{(1;7);(7;1);(-1;-7);(-7;-1)}

=>(a;b)∈{(-2;4);(4;-2);(-4;-10);(-10;-4)}

TH1: (a;b)∈{(-2;4);(4;-2)}

(x+a)(x+b)

=(x-2)(x+4)

\(=x^2+2x-8\)

(x+m)(x-3)+7=(x-2)(x+4)

=>\(x^2-3x+mx-3m+7=x^2+2x-8\)

=>x(m-3)-3m+7=2x-8

=>m-3=2 và -3m+7=-8

=>m=5

TH2: \(\left(a;b\right)\in\left\lbrace\left(-4;-10\right);\left(-10;-4\right)\right\rbrace\)

=>(x+a)(x+b)=(x-4)(x-10)

(x+m)(x-3)+7=(x-4)(x-10)

=>\(x^2-3x+mx-3m+7=x^2-14x+40\)

=>x(m-3)-3m+7=-14x+40

=>m-3=-14 và -3m+7=40

=>m=-11

ta co' 
(x+a).(x-4)-7=(x+b).(x+c) 
nen voi x=4 thi 
-7=(4+b)(4+c)=-7.1=7.(-1) 
do a,c,b∈Z va b,c co vai tro nhu nhau nen gia su b>=c 
co 2 TH xay ra 
**{4+b=7│4+c=-1}↔{b=3│c=-5}suy ra a=2 
ta co(x+2)(x-4_-7=(x+3)(x-5) 
** {4+b=1│4+c=-7}↔{b=-3│c=-11} suy ra a=-10 
ta co(x-10)(x-4)-7=(x-3)(x-11)

x^2+(a-5)x-5a+2=x^2+(b+c)x+bc

=> \(\hept{\begin{cases}a-5=b+c\\2-5a=bc\end{cases}\Leftrightarrow}\)

giải pt nghiệm nguyên=>

a=-2

b=-4

c=-3

a: \(\frac{P\left(x\right)}{2x+3}=\frac{6x^3-7x^2-16x+m}{2x+3}\)

\(=\frac{6x^3+9x^2-16x^2-24x+8x+12+m-12}{2x+3}\)

\(=3x^2-8x+4+\frac{m-12}{2x+3}\)

Để P(x)⋮2x+3 thì m-12=0

=>m=12

b: m=12

=>P(x)=\(6x^3-7x^2-16x+12\)

\(=6x^3-4x^2-3x^2+2x-18x+12\)

=(3x-2)(\(2x^2-x-6\) )

=>P(x) chia 3x-2 dư R=0

c: \(P\left(x\right)=\left(3x-2\right)\left(2x^2-x-6\right)\)

\(=\left(3x-2\right)\left(2x^2-4x+3x-6\right)\)

=(3x-2)(x-2)(2x+3)