ứng dụng t/c thêm bớt nà bạn 

x⁴+x²+1

=(x²)²+2x²+1-2x²+x²

=(x²+1)²-2x²+x² 

= (x² + 1)² − x² 

= (x² + x+ 1 )(x² − x+ 1 )

\(x^4+x^2+1\)
\(=\left[\left(x^2\right)^2+2.x^2.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]-\left(\frac{1}{2}\right)^2+1\)
\(=\left(x^2+\frac{1}{2}\right)^2-\frac{1}{4}+\frac{4}{4}\)
\(=\left(x^2+\frac{1}{2}\right)^2+\frac{3}{4}\)

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

x²-4x+x+4

=x²+x-4x+4

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

=(x+1)(x-4)

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

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

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

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

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

x^2-4+4xy-8y=x^2+4xy+4y^2-4y^2-8y-4=(x+2y)^2-(2y+2)^2=(x+2y-2y+2)(x+2y+2y-2)=(x+2)(x+4y-2)

ta có : x^8 +x^4 +1= (x^8 -x^5) +(x^5-x^2) +(x^4 -x) +(x^2 +x 1)=x^5.(x^3 -1) +x^2(x^3-1) +x(x^3-1) +(x^2 +x+1)=x^5.(x-1)(x^2 +x+1) +x^2(x-1)(x^2 +x+1) +x(x-1)(x^2 +x+1) +(x^2 +x+10=(x^2 +x+1)(x^6- x^5 +x^3 -x +1)

b/ \(=x^8-x^7+x^5-x^4+x^2+x^6-x^5+x^3-x^2+1+x^7-x^6+x^4-x^3+x\)

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

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