Ta có : \(x^4-5x^2+4\)

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

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

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

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

Ta có: \(x^4-5x^2+4\)

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

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

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

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

${}^{}-2{}^{}+2x-1$

$={}^{}-{}^{}-{}^{}+{}^{}-{}^{}+x+x-1$

$={}^{}\left(x-1\right)-{}^{}\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)$

$=\left(x-1\right)\left({}^{}-{}^{}-x+1\right)$
$=\left(x-1\right)[$

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

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

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

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

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

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

Trả lời:

2x³ - 8x² - 8x 

= 2x ( x² - 4x - 4 )

\(\left(2x-2\right)^2\)

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

\(x^2+8x+16=\left(x+4\right)^2\)

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

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

\(x^2-8x+12=\left(x^2-6x\right)-\left(2x-12\right)=x\left(x-6\right)-2\left(x-6\right)=\left(x-2\right)\left(x-6\right)\)

=(x-2)(x-6)

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

\(=\left(2x-2-\sqrt{3}\right)\left(2x-2+\sqrt{3}\right)\)

\(14x^2-14xy-8x+8y=14x\left(x-y\right)-8\left(x-y\right)=\left(x-y\right)\left(14x-8\right)\)

\(14x^2-14xy-8x+8y\)

\(=14x\left(x-y\right)-8\left(x-y\right)\)

\(=\left(14x-8\right)\left(x-y\right)\)

\(=2\left(7x-4\right)\left(x-y\right)\)

\(14x^2-14xy-8y+8y=2\left(x-y\right)\left(7x-4\right)\)