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

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

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

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

3x^3 - 14x^2 + 4x + 3
= (3x^3+x^2) - 15^2- 5x+ 9x+ 3
= x^2(3x+1)- 5x(3x+1)+ 3(3x+1)
= (x^2- 5x+ 3)(3x+1)

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

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)

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

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

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

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

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

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

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

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

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

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

Ta có : 5x² + 14x - 3

= 5x² + 15x - x - 3

= (5x² + 15x) - (x + 3)

= 5x(x + 3) - (x + 3)

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

Phân tích đa thức thành nhân tử 5x2+14x−3

Theo đề bài ta có:

\(5x^2+14x-3\)

\(\Leftrightarrow5x^2+15x-x-3\)

\(\Leftrightarrow\left(5x^2+15x\right)-\left(x-3\right)\)

\(\Leftrightarrow5x\left(x-3\right)-\left(x+3\right)\)

\(\Leftrightarrow\left(3x\right)\left(5x-1\right)\)

\(4x^4+2x^3-8x^2+3x+9\)

\(=4x^4+4x^3-2x^3-2x^2-6x^2-6x+9x+9\)

\(=\left(x+1\right)\left(4x^3-2x^2-6x+9\right)\)

\(=\left(x+1\right)\left(4x^3+6x^2-8x^2-12x+6x+9\right)\)

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

\(x^3-x^2-14x+24\)

\(=x^3-2x^2+x^2-2x-12x+24\)

\(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+x-12\right)\)

\(=\left(x-2\right)\left(x^2+4x-3x-12\right)\)

\(=\left(x-2\right)\left[x\left(x+4\right)-3\left(x+4\right)\right]\)

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

Ta có:\(x^3-x^2-14x+24=\left(x^3-2x^2\right)+\left(x^2-2x\right)-\left(12x-24\right)\)

                                                    \(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)

                                                    \(=\left(x-2\right)\left(x^2+x-12\right)\)

                                                    \(=\left(x-2\right)\left(x^2-3x+4x-12\right)\)

                                                    \(=\left(x-2\right)\left[x\left(x-3\right)+4\left(x-3\right)\right]\)

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

Ta có:

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

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

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

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