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

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

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

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

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

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

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

x^5+2x^4+2x^3+2x^2+2x+1

=(x^5+x^4)+(x^4+x^3)+(x^3+x^2)+(x^2+x)+(x+1)

=x^4(x+1)+x^3(x+1)+x^2(x+1)+x(x+1)+(x+1)

=(x+1)(x^4+x^3+x^2+x+1)

\(x^4+2019x^2+2018x+2019\)

\(=x^4+x^2+1+2018\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2018\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)

\(B=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt:  \(x^2+7x+10=t\)Khi đó B trở thành:

\(B=t\left(t+2\right)-24\)

\(=t^2+2t-24=\left(t-4\right)\left(t+6\right)\)

đến đây bạn thay trở lại

x²-4x+x+4

=x²+x-4x+4

=x(x+1)-4(x+1)

=(x+1)(x-4)

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

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

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

\(=x\left(x+1\right)-4\left(x+1\right)\)

\(=\left(x-4\right)\left(x+1\right)\).

\(3x-7\sqrt{x}+4=3x-4\sqrt{x}-3\sqrt{x}+4=\sqrt{x}\left(3\sqrt{x}-4\right)-\left(3\sqrt{x}-4\right)=\left(3\sqrt{x}-4\right)\left(\sqrt{x}-1\right)\)

\(3x-7\sqrt{x}+4=3x-3\sqrt{x}-4\sqrt{x}+4\)

\(=3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)\)

\(=\left(3\sqrt{x}-4\right)\cdot\left(\sqrt{x}-1\right)\)

\(3x-7\sqrt{x}+4=3x-3\sqrt{x}-4\sqrt{x}+4=3.\left(\sqrt{x}\right)^2-3\sqrt{x}-4\sqrt{x}+4=3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)=\left(3\sqrt{x}-4\right)\left(\sqrt{x}-1\right)\)