a)\(f\left(x\right)=2x^2-x-3+5=\left(x+1\right)\left(2x-3\right)+5\)

Để \(f\left(x\right)⋮g\left(x\right)\Leftrightarrow\left(x+1\right)\left(2x-3\right)+5⋮\left(x+1\right)\)

\(\Leftrightarrow5⋮\left(x+1\right)\)

mà \(x+1\in Z\Rightarrow x+1\in U\left(5\right)=\left\{-1;1;5;-5\right\}\)

\(\Leftrightarrow x\in\left\{-2;0;4;-6\right\}\)

Vậy...

b) \(f\left(x\right)=3x^2-4x+6=\left(3x^2-4x+1\right)+5=\left(3x-1\right)\left(x-1\right)+5\)

Để \(f\left(x\right)⋮g\left(x\right)\Leftrightarrow\left(3x-1\right)\left(x-1\right)+5⋮\left(3x-1\right)\)

\(\Leftrightarrow5⋮\left(3x-1\right)\) mà \(3x-1\in Z\Rightarrow3x-1\in U\left(5\right)=\left\{-1;1;5;-5\right\}\)

\(\Leftrightarrow x\in\left\{0;\dfrac{2}{3};2;-\dfrac{4}{3}\right\}\) mà x nguyên\(\Rightarrow x\in\left\{0;2\right\}\)

Vậy...

c)\(f\left(x\right)=\left(-2x^3-7x^2-5x+2\right)+3\)\(=\left(-2x^3-4x^2-3x^2-6x+x+2\right)+3\)\(=\left[-2x^2\left(x+2\right)-3x\left(x+2\right)+\left(x+2\right)\right]+3\)

\(=\left(x+2\right)\left(-2x^2-3x+1\right)+3\)

Làm tương tự như trên \(\Rightarrow x+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Leftrightarrow x\in\left\{-5;-3;-1;1\right\}\)

Vậy...

d)\(f\left(x\right)=x^3-3x^2-4x+3=x\left(x^2-3x-4\right)+3=x\left(x+1\right)\left(x-4\right)+3\)

Làm tương tự như trên \(\Rightarrow x+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow x\in\left\{-4;-2;0;2\right\}\)

Vậy...

\(a,f\left(x\right):g\left(x\right)=\left(3x^4+9x^3+7x+2\right):\left(x+3\right)\\ =\left[3x^3\left(x+3\right)+7\left(x+3\right)-19\right]:\left(x+3\right)\\ =\left[\left(3x^3+7\right)\left(x+3\right)-19\right]:\left(x+3\right)\\ =3x^3+7.dư.19\)

\(c,\) Để \(k\left(x\right)⋮g\left(x\right)\Leftrightarrow-x^3-5x+2m=\left(x+3\right)\cdot a\left(x\right)\)

Thay \(x=-3\)

\(\Leftrightarrow-\left(-3\right)^3-5\left(-3\right)+2m=0\\ \Leftrightarrow27+15+2m=0\\ \Leftrightarrow2m=-42\\ \Leftrightarrow m=-21\)

Bài 1:

Ta có: \(5x^3-3x^2+2x+a⋮x+1\)

\(\Leftrightarrow5x^3+5x^2-8x^2-8x+10x+10+a-10⋮x+1\)

\(\Leftrightarrow a-10=0\)

hay a=10

b: Ta có: f(x):g(x)

\(=\dfrac{x^3-2x^2+3x+a}{x+1}\)

\(=\dfrac{x^3+x^2-3x^2-3x+6x+6+a-6}{x+1}\)

\(=x^2-3x+6+\dfrac{a-6}{x+1}\)

Để f(x):g(x) là phép chia hết thì a-6=0

hay a=6

a: Thay a=3 vào f(x), ta được:

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

\(\dfrac{f\left(x\right)}{g\left(x\right)}=\dfrac{x^3-2x^2+3x+3}{x+1}\)

\(=\dfrac{x^3+x^2-3x^2-3x+6x+6-3}{x+1}\)

\(=x^2-3x+6-\dfrac{3}{x+1}\)

d: Ta có: f(x):g(x)

\(=\dfrac{x^3-2x^2+3x+5}{x+1}\)

\(=\dfrac{x^3+x^2-3x^2-3x+6x+6-1}{x+1}\)

\(=x^2-3x+6+\dfrac{-1}{x+1}\)

Để f(x) chia hết cho g(x) thì \(x+1\in\left\{1;-1\right\}\)

hay \(x\in\left\{0;-2\right\}\)

Lời giải:
a. $f(x)=x^4-3x^2+2x-7=x^3(x+2)-2x^2(x+2)+x(x+2)-7$

$=(x+2)(x^3-2x^2+x)-7=g(x)(x^3-2x^2+x)-7$

Vậy $f(x)$ chia $g(x)$ được thương là $x^3-2x^2+x$ và dư là $-7$

b. Theo phần a $f(x)=(x^3-2x^2+x)g(x)-7$

Với $x$ nguyên, để $f(x)\vdots g(x)$ thì $7\vdots g(x)$

$\Leftrightarrow x+2$ là ước của $7$

$\Rightarrow x+2\in\left\{\pm 1;\pm 7\right\}$

$\Leftrightarrow x\in\left\{-3; -1; 5; -9\right\}$

c.

Theo định lý Bezout về phép chia đa thức, để $K(x)=-2x^3+x-m\vdots x+2$ thì: $K(-2)=0$

$\Leftrightarrow -2(-2)^3+(-2)-m=0$

$\Leftrightarrow 14-m=0$

$\Leftrightarrow m=14$