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: \(x^4-5x^2+4\)

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

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

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

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

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

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

Đa thức không phân tích được bạn ơi

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

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

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

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

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

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

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

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

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

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

(a+b)³-(a-b)³=a³+3a²b+3ab²+b³-(a³-3a²b+3ab²-b³)

                    =a³+3a²b+3ab²+b³-a³+3a²b-3ab²+b³

                        =6a²b+2b³

Áp dụng hđt a³-b³=(a-b)(a²+ab+b²) ấy

\(\left(a+b\right)^3-\left(a-b\right)^3=\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

\(=\left(a+b-a+b\right)\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)

\(=2b\left(3a^2+b^2\right)\)

Ta có : \(\left(a+b\right)^3-\left(a-b\right)^3\) 

\(=a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\)

\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3\)

\(=6a^2b+2b^3\)

\(x^3-x^2-21x+45\)

\(=\left(x^3-3x^2\right)+\left(2x^2-6x\right)+\left(-15x+45\right)\)

\(=x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)\)

\(=\left(x^2+2x-15\right)\left(x-3\right)\)

\(=\left[\left(x^2-3x\right)+\left(5x-15\right)\right]\left(x-3\right)\)

\(=\left[x\left(x-3\right)+5\left(x-3\right)\right]\left(x-3\right)\)

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

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

Đặt \(x^2+x+1=t\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=t\left(t+1\right)-12=t^2+t-12=\left(t^2+t+\dfrac{1}{4}\right)-\dfrac{49}{4}=\left(t+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(t+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(t+\dfrac{1}{2}+\dfrac{7}{2}\right)=\left(t-3\right)\left(t+4\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

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

= \(\left(x^2+x+1\right)^2\left(x^2+x+1\right)-12\)

= \(\left(x^2+x+1\right)\left(x^2+x+1\right)-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-4.3\)

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

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

\(x^2-4y^2-12y-9\)

\(=x^2-\left(4y^2+12y+9\right)\)

\(=x^2-[\left(2y\right)^2+2.2x.3+3^2]\)

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

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