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.

Nham nghiem thi ta thay pt co 1 ngiem x=2/3

=> tach nhu sau :

\(3x^3-2x^2-3x^2+2x+3x-2\) 

\(3x^2\left(x-\frac{2}{3}\right)-3x\left(x-\frac{2}{3}\right)+3\left(x-\frac{2}{3}\right)\)

\(\left(x-\frac{2}{3}\right)\left(3x^2-3x+3\right)\)

Chuc ban hoc tot

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

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

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

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

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

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

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

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

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

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

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

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

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

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

mk chỉnh lại đề

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

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

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

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

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

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

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

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

\(A=3x^2-22xy-4x+8y+7y^2+1\)

Giả sử:

\(A=\left(3x+ay+b\right)\left(x+cy+d\right)\)

\(=3x^2+3cxy+3dx+axy+acy^2+ady+bx+bcy+bd\)

\(=3x^2+acy^2+\left(3c+a\right)xy+\left(3d+b\right)x+\left(ad+bc\right)y+bd\)

Ta có:

\(\begin{cases}\begin{matrix}ac=-7\\3c+a=-22\\3d+b=-4\\ad+bc=8\end{matrix}\\bd=1\end{cases}\)\(\Rightarrow\begin{cases}a=-1\\b=-1\\c=-7\\d=-1\end{cases}\)

Vậy \(A=\left(3x-y-1\right)\left(x-7y-1\right)\)

Chúc bạn học tốt ^^

Đặt \(P\left(x\right)=2x^4+3x^3-9x^2-3x+2\)

Giả sử nhân tử của P(x) có dạng : \(P\left(x\right)=2\left(x^2+ax+b\right)\left(x^2+cx+d\right)=\left(x^2+ax+b\right)\left(2x^2+2cx+2d\right)\)

Khai triển : \(P\left(x\right)=2x^4+2cx^3+2dx^2+2ax^3+2acx^2+2adx+2bx^2+2bcx+2bd\)

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

Dùng phương pháp hệ số bất định :

\(\Rightarrow\begin{cases}2a+2c=3\\2ac+2b+2d=-9\\2ad+2bc=-3\\bd=1\end{cases}\) . Giải ra được \(\begin{cases}a=-1\\b=-1\\c=\frac{5}{2}\\d=-1\end{cases}\)

Vậy \(P\left(x\right)=2\left(x^2-x-1\right)\left(x^2+\frac{5}{2}x-1\right)=\left(x^2-x-1\right)\left(2x^2+5x-2\right)\)