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

b. 2x³-3x²+3x-1=2x³-x²-2x²+x+2x-1

= x²(2x-1)-x(2x-1)+(2x-1)

=(2x-1)(x²-x-1)

c. 3x³-14x²+4x+3= 3x³+x²-15x²-5x+9x+3

=x²(3x+1)-5x(3x-1)+3(3x+1)

=(3x+1)(x²-5x+3)

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

Bài làm 

3x2 – 8x + 4 = (4x2 – 8x + 4) - x2 = (2x – 2)2 – x2 = (2x – 2 + x)(2x – 2 – x) 
= (x – 2)(3x – 2)

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.

\(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\\ x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\\ 45+x^3-5x^2-9x=x^2\left(x-5\right)-9\left(x-5\right)=\left(x-3\right)\left(x+3\right)\left(x-5\right)\)

c)=3x(x-y)-5(x-y)

   =(3x-5)(x-y)

Bn j ơi gõ latex đi bn !

\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8+\dfrac{3}{2}x\right)^2-\dfrac{1}{4}x^2=\left(x^2+\dfrac{11}{2}x+8\right)^2-\left(\dfrac{1}{2}x\right)^2=\left(x^2+\dfrac{11}{2}x+8-\dfrac{1}{2}x\right)\left(x^2+\dfrac{11}{2}x+8+\dfrac{1}{2}x\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)

\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)+2x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+4x+8\right)\left(x^2+5x+8\right)+2x\left(x^2+5x+8\right)\)

\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)