4x^2 + 8x

= 4x . x + 4x . 2

= 4x ( x + 2 )

\(4x^{^{ }2}+8x\) 

\(=4x.x+4x.2\)

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

\(a,=4x\left(x+2\right)\\ b,=\left(x-3\right)\left(x+3\right)\\ c,=x^2\left(2x-3\right)+\left(2x-3\right)=\left(2x-3\right)\left(x^2+1\right)\)

a)4x²+8x                     b)x²-9

=4x(x+2)                      =x²-3²

                                     =(x-3)(x+3)

c)2x³-3x²+2x-3

=2x³+2x-(3x²+3)

=2x(x²+1)-3(x²+1)

=(2x-3)(x²+1)

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

\(=x^4+16x^2+9+8x^3-24x-6x^2-5x^2-20x+15+6x^2\)

\(=x^4+8x^3+11x^2-44x+24\)

\(=\left(x^4-x^3\right)+\left(9x^3-9x^2\right)+\left(20x^2-20x\right)-\left(24x-24\right)\)

\(=x^3\left(x-1\right)+9x^2\left(x-1\right)+20x\left(x-1\right)-24\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3+9x^2+20x-24\right)\)

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

\(4x^2-9y^2+4x-6y\)

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

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

 a,  \(3x^3-4x^2+5x-4\)

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

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

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

b,   \(4x^3-3x^2+5x-21\)

\(=4x^3-7x^2+4x^2-7x+12x-21\)

\(=x^2\left(4x-7\right)+x\left(4x-7\right)+3\left(4x-7\right)\)

\(=\left(x^2+x+3\right)\left(4x-7\right)\)

c,   \(3x^3+8x^2+14x+15\)

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

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

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

Bài này dùng phương pháp nhẩm nghiệm (tối ưu nhất với đa thức bậc ba)

Chúc bạn học tốt.

\(x^3-3x^2-4x+12\)

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

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

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

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

\(x^3-3x^2-4x+12\)

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

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

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

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

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

\(=\left(x^2+4x+8\right)\left(x^2+6x+8\right)+x\left(x^2+6x+8\right)\)

\(=\left(x^2+4x+8+x\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

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

\(=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

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

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

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

\(=x\left(x-1\right)-3\left(x-1\right)\)

\(=\left(x-1\right)\left(x-3\right)\)

x²-4x+3

=x.x-x-3x-3

=x.(x-1)-3.(x-1)

=(x-1)(x-3)

x² - 4x + 3

= x² - x - 3x + 3 ( dùng phương pháp tách nhân tử )

= ( x² - x ) - ( 3x - 3 ) ( vì đằng trc có đấu " - " nên ta đổi đáu trong ngoặc )

= x( x - 1 ) - 3(x - 1 ) ( đặt nhân tử chung )

= ( x - 1 ) ( x - 3 )

hok tốt !

\(\left(x^2+x\right)^2+4x^2+4x-12=\left[\left(x^2+x\right)^2+4\left(x^2+x\right)+4\right]-16=\left(x^2+x+2\right)-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

\(\left(x^2+x\right)^2+4x^2+4x-12\\ =\left(x^2+x+2\right)-4\\ =\left(x^2+x-2\right)\left(x^2+x+6\right)\)

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

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

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