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

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

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

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

\(b,x^4+5x^3+10x-4=\left(x^4-4\right)+\left(5x^3-10x\right)\)\(=\left(x^2+2\right)\left(x^2-2\right)+5x\left(x^2+2\right)\)

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

\(=6x^2+5x-3xy\)

\(=x\left(6x+5-3y\right)\)

x² - 5x + 5

= x² - 5(x + 1)

= x² - \(\left[\sqrt{5\left(x+1\right)}\right]^2\)

= [x - \(\sqrt{5\left(x+1\right)}\)][x + \(\sqrt{5\left(x+1\right)}\)]

\(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20=\left[\left(x-1\right)\left(x-7\right)\right].\left[\left(x-3\right)\left(x-5\right)\right]-20\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)

Đặt \(x^2-8x+11=t\) \(\Rightarrow\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20=\left(t-4\right)\left(t+4\right)-20=t^2-16-20=t^2-36=\left(t-6\right)\left(t+6\right)\)\(\Rightarrow\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20=\left(x^2-8x+11-6\right)\left(x^2-8x+11+6\right)=\left(x^2-8x+17\right)\left(x^2-8x+5\right)\)

$ 2x^3 - x^2 + 5x + 3 \\ = 2x^3 + x^2 - 2x^2 - x + 6x + 3 \\ = x^2(2x + 1) - x(2x + 1) + 3(2x + 1) \\ = (2x + 1)(x^2 - x + 3) $

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

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

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

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

Vì \(x^2-x+3=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}+3>0\)

Nên ​​

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

5x^2 + 5xy - x - y 

=5x.(x+y)-(x+y)

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

7x - 6x^2 - 2

=-6x²+3x+4x-2

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

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