\(A=3x^2-14x^2+4x+3\)

Giả sử:

\(A=\left(3x+a\right)\left(x^2+bx+c\right)\)

\(=3x^3+3bx^2+3cx+ax^{2\:}+abx+ac\)

\(=3x^3+\left(3b+a\right)x^2+\left(3c+ab\right)x+ac\)

Ta có:

\(\begin{cases}3b+a=-14\\3c+ab=4\\ac=3\end{cases}\)\(\Rightarrow\begin{cases}a=1\\b=-5\\c=3\end{cases}\)

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

3x^3+x^2 -15x^2-5x+9x+3

= (3x+1)(x^2-5x+3)

\(3x^3-14x^2+4x+3\)

\(=3x^3+x^2-15x^2-5x+9x+3\)

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

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

\(3x^3+5x^2-14x+4\\ =\left(3x^3-x^2\right)+\left(6x^2-2x\right)-\left(12x-4\right)\\ =x^2\left(3x-1\right)+2x\left(3x-1\right)-4\left(3x-1\right)\\ =\left(x^2+2-4\right)\left(3x-1\right)\)

\(4x^3+14x^2+6x\)

\(=2x\left(2x^2+7x+3\right)\)

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

\(=2x\left[2x\left(x+3\right)+\left(x+3\right)\right]\)

\(=2x\left[\left(2x+1\right)\left(x+3\right)\right]\)

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

a) \(-4x^2+10xy-14x\)

\(=2x\left(-2x+5y-7x\right)\)

b) \(3x\left(x-1\right)-y\left(1-x\right)+2x-2\)

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

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

Bài làm

a) -4x² + 10xy - 14x

= 2x( -2x + 5y - 7 )

b) 3x( x - 1 ) - y ( 1 - x ) + 2x - 2

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

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

# Học tốt #

Ta có:

\(3x^3+10x^2+14x+8\)

\(=3x^3+4x^2+6x^2+8x+6x+8\)

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

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