x³ - 9x² + 6x + 16

= x³ - 8x² -x² + 8x - 2x + 16

= x²(x-8) -x(x-8) -2(x-8)

= (x-8)(x²-x-2)

= (x-8)(x²-2x + x - 2)

=(x-8)[x(x-2)+(x-2)]

=(x-8)(x-2)(x+1)

Đặt H \(=x^4-5x^3+7x^2-6\)

Gỉa sử : \(H=\left(x^2+ax+b\right)\left(x^2+cx+d\right)\)

                   \(=x^4+cx^3+dx^2+ax^{3\:}+acx^2+adx+bx^2+bcx+bd\)

                      \(=x^4+\left(a+c\right)x^3+\left(ac+b+d\right)x^2+\left(ad+bc\right)x+bd\)

       \(\Leftrightarrow\hept{\begin{cases}a+c=-5\\ac+b+d=7\\ad+bc=0\end{cases}}\)

                 \(\left\{bd=6\right\}\)

           \(\Leftrightarrow\hept{\begin{cases}a=-3\\b=3\\c=-2\end{cases}}\)

                   \(\left\{d=-2\right\}\)

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

Chúc bạn học tốt !!!

Bó tay với hệ số bất định 

\(3x^4-5x^3-18x^2-3x+5\)

\(=3x^4-6x^3+x^3-15x^2-2x^2-x^2-5x+2x+5\)\(=\left(3x^4-6x^3-15x^2\right)+\left(x^3-2x^2-5x\right)-\left(x^2-2x-5\right)\)

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

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

3x^2+5x -2

=3x^2 -x +6x -2

=(3x^2 -x) + (6x - 2)

=x (3x -1 )+ 2(3x -1)

=(3x - 1)(x+2)

Chúc bạn học tốt!!!!!!!!!!!!! nha.

\(x^4-5x^3+7x^2-6=\left(x^2+ax+b\right)\left(x^2+cx+d\right)\)

\(\Leftrightarrow x^4-5x^3+7x^2-6=x^4+cx^3+dx^2+ax^3+acx^2+adx+bx^2+bcx+db\)

\(\Leftrightarrow x^4-5x^3+7x^2-6=x^4-x^3\left(-c-a\right)+x^2\left(d+ac+b\right)+x\left(ad+bc\right)-\left(-bd\right)\)

Đồng nhất hệ số , ta có :

\(\left\{{}\begin{matrix}-a-c=5\\d+ac+b=7\\ad+bc=0\\-bd=6\end{matrix}\right.\)

\(db=-6\) mà \(bd\in Z\) ...đặt từng gtri rồi giải là Ok .

Đặt: \(H=\)\(x^4-5x^3+7x^2-6\)

Giả sử: \(H=\left(x^2+ax+b\right)\left(x^2+cx+d\right)\)

\(=x^4+cx^3+dx^2+ax^3+acx^2+adx+bx^2+bcx+bd\)

\(=x^4+\left(a+c\right)x^3+\left(ac+b+d\right)x^2+\left(ad+bc\right)x+bd\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+c=-5\\ac+b+d=7\\ad+bc=0\\bd=-6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=3\\c=-2\\d=-2\end{matrix}\right.\)

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

ta có : \(x^4-5x^3+7x^2-6\)

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

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

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