= \(5x^3-2x^2-10x^2+4x+10x-4\)

= \(5x^2\left(x-\frac{2}{5}\right)-10x\left(x-\frac{2}{5}\right)+10\left(x-\frac{2}{5}\right)\)

=\(\left(x-\frac{2}{5}\right)\left(5x^2-10x+10\right)\)

Chuc ban hoc tot!

\(=14x^2-2x+35x-5\)

\(=2x\left(7x-1\right)+5\left(7x-1\right)\)

\(\left(7x-1\right)\left(2x+5\right)\)

Phân tích đa thức thành nhân tử: 14x^2 + 33x - 5

=14x2−2x+35x−5

=2x(7x−1)+5(7x−1)

(7x−1)(2x+5)

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

       \(x^3-9x^2+14x\)

= \(x^3-7x^2-2x^2+14x\)

= \(x^2.\left(x-7\right)-2x.\left(x-7\right)\)

= \(\left(x-7\right).\left(x^2-2x\right)\)

= \(\left(x-7\right).\left(x-2\right).x\)

\(x^3-9x^2+14x\)

\(=x\left(x^2-9x+14\right)\)

\(=x\left(x^2-7x-2x+14\right)\)

\(=x\left[x\left(x-7\right)-2\left(x-7\right)\right]\)

\(=x\left(x-2\right)\left(x-7\right)\)

Bài làm

   3x² + 14x - 15

= 3x² + 9x + 5x - 15

= -( 9x - 3x² ) - ( 15 - 5x )

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

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

# Hokc tốt #

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

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